Saturday

April 19, 2014

April 19, 2014

Number of results: 224,839

**Precalculus with Trigonometry**

For the first problem: cos(-x) - sin(-x) = cos(x) + sin(x) A few identities for negatives: cos(-x) = cos(x) sin(-x) = -sin(x) Therefore: cos(x) - [-sin(x)] = cos(x) + sin(x) ================================== For your last problem: cos(x+y)cos(x-y) = cos^2(x) - sin^2(y) Some ...
*Thursday, April 4, 2013 at 6:11pm by MathGuru*

**Precalculus**

making everything sin and cos, we have (cos^2/sin)/(1/cos + sin/cos) + sin/(1/cos - sin/cos) cos^2/sin * cos/(1+sin) + sin*cos/(1-sin) cos^3/(sin(1+sin)) + sin*cos/(1-sin) (cos^3(1-sin) + sin^2*cos(1+sin))/(sin(1-sin^2)) (cos^3 - sin*cos^3 + cos*sin^2 + sin^3)/(sin*cos^2) (cos...
*Tuesday, October 23, 2012 at 8:55pm by Steve*

**math**

(cos/sin/(1 - sin/cos) + sin/cos/(1-cos/sin) - sin/cos - cos /sin multiply top and bottom of all by sin cos cos^2/(sin cos -sin^2) + sin^2/(sin cos - cos^2) - (sin^2+cos^2)/sin cos cos^2/(sin cos -sin^2) + sin^2/(sin cos - cos^2) - 1 /sin cos cos^2/(sin (cos -sin)) - sin^2/(...
*Tuesday, February 10, 2009 at 6:19pm by Damon*

**PreCalculus (PLEASE HELP, IM BEGGING!)**

cos/(1-cos) * ((1+sin)+sin)/(1+sin) cos(1+2sin)/((1-cos)(1+sin)) Not sure how to make it any simpler. No chance of a typo in the problem? ---------------------------- ((1+cos) + (1-cos))/(1-cos)(1+cos) 2/sin^2 ---------------------- (1-cos^2/sin^2)/(1+cos^2/sin^2) + 1 (sin^2...
*Saturday, May 5, 2012 at 2:09am by Steve*

**Pre-Calculus**

cos^4 - sin^4 = (cos^2+sin^2)(cos^2-sin^2) Remember algebra I and your double-angle formulas? working with the right side, we have csc^2 - csc cot 1/sin^2 (1 - cos) 1/(1-cos^2) (1-cos) (1-cos) / (1+cos)(1-cos) 1/(1+cos) Hmmm. I don't get 1/cos (cos x/ (sec x -1))- (cos x/ tan^...
*Thursday, October 31, 2013 at 10:03am by Steve*

**Math**

a good place to start is usually to make everything sin/cos: sin^2 + sin^2/cos^2 [(1-cos^2)cos^2 + sin^2]/cos^2 (cos^2 - cos^4 + 1 - cos^2)/cos^2 (1 - cos^4)/cos^2 1/cos^2 - cos^2 sec^2 - cos^2 _______________________ sin^2/cos^2 + sin^2 tan^2 + sin^2
*Thursday, November 17, 2011 at 4:40pm by Steve*

**Math**

often it's easier to work with just sin and cos. working just on the left side, we have sin/(cot+1) + cos/(tan+1) sin/(cos/sin+1) + cos/(sin/cos+1) sin^2/(cos+sin) + cos^2(sin+cos) (sin^2 + cos^1)/(sin+cos) 1/(sin+cos) ta-daaaah
*Thursday, November 1, 2012 at 10:37pm by Steve*

**trigonometry**

tan(sin+cot*cos)/cos When dealing with such a combination of functions, it is usually good to start out by converting everything to sin and cos: (sin/cos)(sin+(cos/sin)cos)/cos sin/cos * (sin^2 + cos^2)/sin * 1/cos (sin^2 + cos^2)/cos^2 1/cos^2 sec^2
*Friday, May 4, 2012 at 4:30am by Steve*

**tigonometry**

expres the following as sums and differences of sines or cosines cos8t * sin2t sin(a+b) = sin(a)cos(b) + cos(a)sin(b) replacing by by -b and using that cos(-b)= cos(b) sin(-b)= -sin(b) gives: sin(a-b) = sin(a)cos(b) - cos(a)sin(b) Add the two equations: sin(a+b) + sin(a-b) = ...
*Sunday, November 26, 2006 at 6:35pm by Pablo*

**Trigonometry**

well, i think u have mistaken my question. what i meant is i cant memorize the sum-to-product formula and product to sum formula, which are Review Product to Sum Formulas 1. sin x cos y = (1/2) [ sin(x + y) + sin(x - y) ] 2. cos x sin y = (1/2) [ sin(x + y) - sin(x - y) ] 3. ...
*Monday, July 21, 2008 at 7:07am by Tommy*

**trignometry**

cos 2x = cos(x+x) = cosxcosx - sinxsinx = cos^2 x - sin^2 x = cos^2 x - (1 - cos^2 x) = 2 cos^2 x - 1, the last part of your equation. all we need it to look at now is the middle part of (1 - tan^2 x)/(1 + tan^2 x) , notice the necessary brackets = (1 - sin^2 x/cos^2 x)/(1 + ...
*Saturday, October 6, 2012 at 12:20pm by Reiny*

**trigonometry**

changing all to sinx and cosx, we have on the left: (cos/sin - 1)/(cos/sin + 1) = (cos-sin)/sin / (cos+sin)/sin = (cos-sin)/(cos+sin) = (cos-sin)^2 / (cos^2-sin^2) = (cos^2 - 2sin*cos + sin^2)/(cos^2-sin^2) = (1-sin2x)/cos2x
*Sunday, December 9, 2012 at 5:33pm by Steve*

**Trigonometry**

2) csc(x)=1/sin(x) and cot(x)=cos(x)/sin(x) => cot^3(x)/csc(x)=cos^3(x)*sin(x)/sin^3(x) =cos^3(x)/sin²(x) Now, we know that cos²(x)=1-sin²(x) => cos^3(x)=cos(x)*(1-sin²(x))=cos(x)-cos(x)*sin²(x) So, when we fill this in in the equation we get that: cos^3(x)/sin²(x)=(cos(...
*Monday, February 25, 2008 at 7:36pm by Christiaan*

**precalc h**

maybe you mean sin t/(1-cos t) + (1-cos t)/sin t sin^2 t (1-cos t) (1-cos t)^2 ----------------- + ---------------- (1-cos t)(sin t) ______ (1-cos t)(sin t) numerator only for a while sin^2 t-sin^2 t cos t +1 -2 cos t+cos^2 t but sin^2 + cos^2 = 1 2 -sin^2 t cos t - 2 cos t ...
*Sunday, December 9, 2012 at 8:17pm by Damon*

**algebra**

Can someone please help me do this problem? That would be great! Simplify the expression: sin theta + cos theta * cot theta I'll use A for theta. Cot A = sin A / cos A Therefore: sin A + (cos A * sin A / cos A) = sin A + sin A = 2 sin A I hope this will help. in my algebra ...
*Sunday, February 18, 2007 at 9:01am by Valerie*

**Math - Calculus**

The identity below is significant because it relates 3 different kinds of products: a cross product and a dot product of 2 vectors on the left side, and the product of 2 real numbers on the right side. Prove the identity below. | a × b |² + (a • b)² = |a|²|b|² My work, LSH...
*Monday, March 3, 2008 at 3:17pm by Anonymous*

**Calculus**

sin (a+b) = sin a cos b + cos a sin b cos (a+b) = cos a cos b - sin a sin b multiply the first by cos b multiply the second by -sin b then add them and you get sin a cos^2 b + cos a sin b cos b -cos a cos b sin b + sin a sin^2 b low and behold that is sin a [ cos^2 b + sin^2 b...
*Monday, January 7, 2008 at 2:22pm by Damon*

**Trig please help!**

LS looks like the sum of cubes sin^6 x + cos^6 x = (sin^2x)^3 + (cos^2x)^3 = (sin^2x+cos^2x)(sin^4x - (sin^2x)(cos^2x) + sin^4x) = (1)(sin^4x - (sin^2x)(cos^2x) + sin^4x) Now let's do some "aside" (sin^2x + cos^2)^2 would be sin^4x + 2(sin^2x)(cos^2x) + cos^4x we 'almost' have...
*Monday, February 23, 2009 at 4:05pm by Reiny*

**Trig!**

Sounds like a good justification to me. Oh, did you mean prove them? In that case, using the identities, cos(a-b) = cos(a + (-b)) = cos(a) cos(-b) - sin(a) sin(-b) = cos(a)cos(b) + sin(a) sin(b) sin(a-b) = sin(a + (-b)) = sin(a) cos(-b) + cos(a) sin(-b) = sin(a) cos(b) - cos(a...
*Tuesday, September 27, 2011 at 12:21pm by Steve*

**math**

a^3=1/cos-cos = (1-cos^2)/cos=sin^2/cos b^3=1/sin-sin = (1-sin^2)/sin=cos^2/sin a^3/b^3=sin^2/cos * cos^2/sin=sin^3/cos^3 => a/b=sin/cos = tan (Hope you understand your own notations.)
*Sunday, May 19, 2013 at 8:59am by MathMate*

**Math**

OK, but the effort is the same. cos/(1+sin) Now, you know that a^2-b^2 = (1+a)(1-a), and you knwo that cos^2 = 1-sin^2, so that should lead you to simplify the denominator by multiplying top and bottom by (1-sin) cos/(1+sin) * (1-sin)/(1-sin) = cos(1-sin)/(1-sin^2) = cos(1-sin...
*Saturday, February 23, 2013 at 7:03pm by Steve*

**precalc h**

1+1/cos = tan^2/sec-1 I will guess (parentheses missing again) you mean 1+1/cos = tan^2/(sec-1 ) use sin and cos everywhere 1+1/cos = (sin/cos)^2/([1/cos]-1 ) multiply top and bottom of right by cos^2 1 + 1/cos = sin^2/(cos - cos^2) (cos + 1)/cos = sin^2/cos(1-cos) (cos + 1)/...
*Sunday, December 9, 2012 at 8:17pm by Damon*

**math**

It will surprise you to see how easy this one is. we know that sin^2 a + cos^2 a = 1 and sin^2 b + cos^2 b = 1 then sin^2 a + cos^2 a = sin^2 b + cos^2 b now let's "move" the terms cos^2 a and cos^2 b to get the needed equation sin^2 a - cos^2 b = sin^2 b - cos^2 a Done!
*Monday, April 6, 2009 at 3:46pm by Reiny*

**Math**

since there's only one side that needs manipulating, I'd pick the left side. :-) since 1+tan = (sin+cos)/cos and 1-cot = (sin-cos)/sin we have [(sin+cos)*cos/(sin+cos)]^2 + [(sin-cos)*sin/(sin-cos)]^2 = sin^2+cos^2 = 1
*Friday, November 2, 2012 at 12:33pm by Steve*

**Trig.**

sec^2(x)cot(x) - cot(x) = tan(x) Convert everything to sine and cosine using the identity tan(x) = sin(x)/cos(x). cos^-2(x)(cos(x)/sin(x)) - cos(x)/sin(x) = sin(x)/cos(x) 1/( cos(x)sin(x) ) - cos(x)/sin(x) = sin(x)/cos(x) Note: it is important to write sin(x) as opposed to sin...
*Saturday, January 9, 2010 at 10:27am by Marth*

**Pre-Calc/Math**

One way to deal with identities is to convert every term to sin and cos: (sec x/csc x) + (sin x/cos x) =(1/cos (x))÷ (1/sin(x)) + sin(x)/cos(x) =sin(x)/cos(x)+sin(x)/cos(x) =2tan(x) =2/cot(x)
*Monday, May 14, 2012 at 7:13pm by MathMate*

**math**

Can you please check my work. A particle is moving with the given data. Find the position of the particle. a(t) = cos(t) + sin(t) s(0) = 2 v(0) = 6 a(t) = cos(t) + sin(t) v(t) = sin(t) - cos(t) + C s(t) = -cos(t) - sin(t) + Cx + D 6 = v(0) = sin(0) -cos(0) + C C=7 2= s(0) = -...
*Thursday, June 23, 2011 at 8:08pm by Kay*

**math**

Your equations contain only trig operators but they have no arguments e.g. sin^2 is meaningless, it has to be something like sin^2Ø or sin^2x Your expression is as meaningless as trying to evaluate 5 + √ anyway ..... for sin^2+tan^2=sec^2-cos^2 LS = sin^2x + sin^2x/cos^...
*Friday, November 18, 2011 at 1:04am by Reiny*

**Trigonometry**

sin^6 θ+cos^6 θ = (sin^2 Ø)^3 + (cos^2 Ø)^3 ---- the sum of cubes = (sin^2 Ø + cos^2 Ø)( (sin^2 Ø)^2 - (sin^2 Ø)(cos^2 Ø) + (cos^2 Ø)^2 ) = (1)( (sin^2 Ø)^2 - (sin^2 Ø)(cos^2 Ø) + (cos^2 Ø)^2 ) now, if sin 2Ø = 1/3 2sinØcosØ = 2/3 sinØcosØ = 1/3 and (sin^2 Ø)^2 + (...
*Monday, May 27, 2013 at 11:51am by Reiny*

**math pre-calc**

One way -- do the algebra involved in the following: sin 5x = sin 3x cos 2x + sin 2x cos 3x where sin 3x = sin 2x cos x + cos 2x sin x cos 3x = cos 2x cos x - sin 2x sin x cos 2x = 1 - 2 sin^2 x sin 2x = 2 sin x cos x and cos^2 x = 1 - sin^2 x substitute these up the line, ...
*Saturday, January 19, 2013 at 9:05pm by bobpursley*

**trig**

Reduce the following to the sine or cosine of one angle: (i) sin145*cos75 - cos145*sin75 (ii) cos35*cos15 - sin35*sin15 Use the formulae: sin(a+b)= sin(a) cos(b) + cos(a)sin(b) and cos(a+b)= cos(a)cos(b) - sin(a)sin)(b) (1)The quantity = sin(145-75) = sin 70 = cos 20 note that...
*Sunday, November 26, 2006 at 6:30pm by Paul C*

**Pre-calc**

on the LS, sin^3+cos^3 = (sin+cos)(sin^2 - sin*cos + cos^2) = (sin+cos)(1-sin*cos) 1-2cos^2 = sin^2-cos^2 = (sin+cos)(sin-cos) divide, giving (1-sin*cos)/(sin-cos) on the RS, sec-sin = 1/cos - sin = (1-sin*cos)/cos tan-1 = sin/cos - 1 = (sin-cos)/cos divide, giving (1-sin*cos...
*Thursday, May 10, 2012 at 11:22pm by Steve*

**Pre-Calc**

on the LS, sin^3+cos^3 = (sin+cos)(sin^2 - sin*cos + cos^2) = (sin+cos)(1-sin*cos) 1-2cos^2 = sin^2-cos^2 = (sin+cos)(sin-cos) divide, giving (1-sin*cos)/(sin-cos) on the RS, sec-sin = 1/cos - sin = (1-sin*cos)/cos tan-1 = sin/cos - 1 = (sin-cos)/cos divide, giving (1-sin*cos...
*Thursday, May 10, 2012 at 11:07pm by Steve*

**Math please help quick**

Hey, there is a limit ! I will do the first one, then you try the rest. Which of the following are identities? Check all that apply. (Points : 2) sin2x = 1 - cos2x I assume you mean sin^2 x not sin 2x This is the same as sin^2 x + cos^2 x = 1 which IS an identity. sin^2x - cos...
*Sunday, January 12, 2014 at 6:11pm by Damon*

**maths**

sin^3 - cos^3 = (sin-cos)(sin^2 + sin*cos + cos^2) = (sin-cos)(1+sin*cos) so, (sin^3-cos^3)/(sin-cos) = 1+sin*cos 1+cot^2 = csc^2, so cos/csc = sin*cos tan*cot=1, so we have 1+sin*cos - sin*cos - 2 = -1 which is true for all values of x
*Wednesday, October 17, 2012 at 12:19pm by Steve*

**Trig**

(sin/cos)(1-cos)/cos sin(1 - cos)/cos^2 I see nothing special about this. could it be: (sin/cos) [1/(cos-1)] ??? then (sin/cos) [1/(cos-1)][(cos+1)/cos+1)] (sin/cos)(cos+1)/(cos^2-1) -(sin/cos)(cos+1)/sin^2 - (cos+1)/sin cos
*Sunday, January 8, 2012 at 7:49pm by Damon*

**precalculus**

my second line would have been (sin e/cos e + cos e)/(1/cos e + cos e/sin e) = [(sin e + cos^2 e)/cos e]/[(sin e + cos^2 e)/(sin e cos e)] = sin e in my last step I inverted and multiplied so the sin e + cos^2 e canceled and the cos e canceled
*Sunday, February 22, 2009 at 2:06pm by Reiny*

**Math**

I am willing to do one just to show you how. Express all functions as sin and cos, for example tan = sin/cos and then remember that sin^2 + cos^2 = 1 4)Which expression is not equivalent to 1? A)sin^2theta+cot^2thetasin^2theta B)sin^2theta/1-costheta -1 C)sec^2theta+tan^2theta...
*Wednesday, February 13, 2008 at 4:15pm by Damon*

**Trigonometry**

multiply top and bottom of the left most by (1-cos x) multiply top and bottom of the second term on the left by (1 + cos x) that gives you the common denominator on the left of (1 - cos^2 x) which is sin^2 x then do the two multiplications on the top to get on the top sin x ...
*Tuesday, December 4, 2007 at 8:35pm by Damon*

**trigo math**

tan*sin+cos = sin^2/cos + cos = (sin^2+cos^2)/cos = 1/cos = sec (tan*cos^2 + sin^2)/sin = (sin*cos + sin^2)/sin = sin(cos+sin)/sin = cos+sin I think you mean (1+tan)/(1-tan) = (1+tan)^2/(1-tan^2) = (1+2tan+tan^2)/(1-tan^2) = (sec^2+2tan)/(1-tan^2) the last one needs some ...
*Thursday, December 13, 2012 at 10:58am by Steve*

**Mathematics - Trigonometric Identities**

Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y)(sin^2y) / (sin^...
*Thursday, November 8, 2007 at 6:17pm by Anonymous*

**trig**

You have a typo which makes it kind of confusing. tan x = n tan y so sin x/cos x = n sin y/cos y n = (cos y/cos x)(sin x/sin y) but m = (sin x/sin y) sin y = (1/m) sin x cos^2 y = 1-sin^2y = 1-(1/m^2) sin^2x so n = (cos y / cos x)m n^2 = (m^2)(cos^2 y/cos^2 x) n^2 = (m^2/cos^...
*Wednesday, October 12, 2011 at 5:03am by Damon*

**math**

cot x - tan x = 2 cot 2x I will be solving the left side and make it look like the right side. Note that cot x = cos(x) / sin(x) and tan x = sin(x) / cos(x): cos(x) / sin(x) - sin(x) / cos(x) Combining, cos^2 (x) - sin^2 (x) / cos(x)*sin(x): Note that the numerator cos^2 (x...
*Monday, November 18, 2013 at 2:42am by Jai*

**Trig**

I bet you mean tan/(sec - 1) = sin/(1-cos) PARENTHESES ARE VITAL otherwise you are just wasting time (sin/cos)/[(1/cos)-cos/cos] = sin/(1-cos) sin/(1-cos) = sin/(1-cos)
*Sunday, January 8, 2012 at 7:49pm by Damon*

**TRIG!**

Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 x = (sin^2x)^3 + (cos^2x)^3 = (sin^2x+cos...
*Monday, February 23, 2009 at 7:58pm by hayden*

**Pre Calculus**

working with the left side, cos/sin * 1/sin + 1/cos cos/sin^2 + 1/cos (cos^2+sin^2)/(sin^2 cos) 1/(sin^2 cos) 1/sin^2 * 1/cos csc^2 sec
*Sunday, November 4, 2012 at 1:17pm by Steve*

**trig**

sin(4x) can be expanded in the usual way by using the doubling formulas: sin(4x) = 2 sin(2x) cos(2x) = 4 sin(x) cos(x)cos(2x)= 4 sin(x) cos(x) [2 cos^2(x) - 1] The general method also works, but it slightly more laborious in this simple case: You start with exp(ix) = cos(x) + ...
*Saturday, July 18, 2009 at 1:20am by Count Iblis*

**Trig**

For the first, I am going to start on the left LS = sin^2 x/cos^2 x - sin^2 x = (sin^2 x - sin^2 xcos^2 x)/sin^2 x = sin^2x(1 - cos^2 x)/sin^2 x = (sin^2 x)(sin^2 x)/cos^2 x = tan^2 x sin^2 x = RS for the second LS = 1/cos^2 x - (sinx)/cosx)]/[cosx/sinx)] = 1/cos^2 x - sin^2 x...
*Thursday, February 28, 2008 at 5:01pm by Reiny*

**trig**

Use the identity: sin(a+b)=sin(a)cos(b)+cos(a)sin(b) Good to know also: sin(a-b)=sin(a)cos(b)-cos(a)sin(b) cos(a+b)=cos(a)cos(b)-sin(a)sin(b) cos(a-b)=cos(a)cos(b)+sin(a)sin(b)
*Tuesday, December 14, 2010 at 8:45am by MathMate*

**math**

I(n) = Integral of dx/cos^n(x) 1/cos^n(x) = cos^2(x)/cos^(n+2)(x) = [1-sin^2(x)]/cos^(n+2)(x). So, I(n) = I(n+2) - Integral of sin^(2)(x)/cos^(n+2)(x) dx Integral of sin^(2)(x)/cos^(n+2)(x) dx = -Integral of sin(x)/cos^(n+2)(x)dcos(x) = (Partial integration) 1/(n+1) sin(x)/cos...
*Thursday, November 8, 2007 at 4:08pm by Count Iblis*

**trig**

check your typing, since I can prove that your Left Side = -1 L.S. = [(1–sin²x)/sin²x]–[(csc²x–1)/cos²x] = cos²x/sin²x - cot²x/cos²x = cos²x/sin²x - 1/sin²x = (cos²x - 1)/sin²x = -sin²x/sin²x = -1
*Sunday, January 24, 2010 at 5:42pm by Reiny*

**error fx**

sin t/(1-cos t) + (1-cos t)/sin t sin^2 t ______________ (1-cos t)^2 ----------------- + ---------------- (1-cos t)(sin t) ______ (1-cos t)(sin t) sin^2 t + 1 -2 cos t +cos^2 t -------------------------------- (1-cos t)(sin t) 2 - 2 cos t -------------- (1-cos t) sin t = 2/sin...
*Sunday, December 9, 2012 at 8:17pm by Damon*

**precalculus**

To make it easier to type, let x = theta/2 1 - 2sin^2 (x) = 2cos^2 (x) - 1 1 - sin^2 (x) - sin^2 (x) = cos^2 (x) + cos^2 (x) - 1 Because sin^2 (x) + cos^2 (x) = 1: 1 - sin^2 (x) = cos^2 (x) 1 - cos^2 (x) = sin^2 (x) cos^2 (x) - sin^2 (x) = cos^2 (x) - sin^2 (x)
*Wednesday, September 23, 2009 at 12:33pm by Marth*

**math**

cos ^ 2 ( x ) - sin ^ 2 ( x ) = cos ( 2x ) 2 sin ( x ) cos ( x ) = sin ( 2 x ) Equation : cos ^ 2 ( x ) - 2 sin( x ) cos( x ) - sin ^ 2 ( x ) = 0 we can write like : cos ( 2 x ) - sin ( 2 x ) = 0 Add sin ( 2 x ) to both sides cos ( 2 x ) - sin ( 2 x ) + sin ( 2 x ) = 0 + sin...
*Saturday, January 12, 2013 at 11:14pm by Bosnian*

**trigonometry**

copy again sin a cos b + cos a sin b = sin (a+b) so sin (7t) 1/(1/cos x -1) - 1/(cos x +1) cos x/(1 - cos x) - cos x/(1+cos x) [cos x(1+cos x) - cos x(1-cos x)]/(1-cos^2 x) [ cos x + cos^2x - cos x +cos^2 x]/sin^2x 2 cos^2 x/sin^2 x 2 cot^2 x
*Tuesday, December 20, 2011 at 8:40pm by Damon*

**Trig: use parentheses**

Verify: (csc(x)+sec(x))/(sin(x)+cos(x))=cot(x)+tan(x) Left hand side (csc(x)+sec(x))/(sin(x)+cos(x)) =(1/sin(x)+1/(cos(x))/(sin(x)+cos(x)) =((cos(x)+sin(x))/(sin(x)cos(x))/(sin(x)+cos(x)) =1/(sin(x)cos(x)) Right hand side: cot(x)+tan(x) =cos(x)/sin(x) + sin(x)/cos(x) =(cos&...
*Wednesday, June 2, 2010 at 6:10pm by MathMate*

**trigonometry**

This time start from the right-hand side by taking advantage of the term cos(x)-sin(x): (cos(x)+sin(x))/(cos(x)-sin(x)) multiply top and bottom by cos(x)+sin(x) (cos(x)+sin(x))^sup2;/(cos²(x)-sin²(x) =(cos²(x)+sin²(x)+2sin(x)cos(x))/(cos²(x)-sin²(...
*Monday, January 4, 2010 at 6:25pm by MathMate*

**Tyler**

tan 2x = sin 2x / cos 2x = 2 sin x cos x /(cos^2 x - sin^2 x) = 2 cos x [ sin x / (cos^2 x-sin^2 x) ] now if we can find where [ sin x / (cos^2x-sin^2x) ] is 1 then we have it well cos^2 x = 1 - sin^2 x so we have sin x /[ 1 - sin^2 x - sin^2 x] = 1 or sin x = 1 - 2 sin^2 x 2 ...
*Saturday, November 29, 2008 at 6:04pm by Damon*

**trig**

Since you did not use parentheses for your numerators and denominators, I spent most of my time figuring out what you probably mean (1 + sin x)/( cos x) + cos x/(1+sin x) = 2/cos x common denominator on left of (1+sin x) cos x [(1 +sin x)^2 +cos^2 x ]/common denom [ 1 + 2 sin ...
*Tuesday, December 4, 2007 at 9:35pm by Damon*

**math**

Prove that for all real values of a, b, t (theta): (a * cos t + b * sin t)^2 <= a^2 + b^2 I will be happy to critique your work. Start on the left, square it, (a * cos t + b * sin t)^2 = a^2 (1 - sin^2t) + 2ab sin t cost+ b^2 (1 - cos^2 t)= a^2 + b^2 - (a sin t - b cos t)^2...
*Tuesday, December 19, 2006 at 11:33pm by mathstudent*

**Trig**

LS = (sin^2 t + cos^2 t)(sin^2 t - cos^2 t)/(sin^2 t cos^2 t) = 1(sin^2 t - cos^2 t)/(sin^2 t cos^2 t) = sin^2 t/(sin^2 t cos^2 t) - cos^2 t/(sin^2 t cos^2 t) = 1/cos^2 t - 1/sin^2 t = sec^2 t - csc^2 t = RS
*Thursday, April 19, 2012 at 8:26pm by Reiny*

**TRIG**

sin (a+b) = sin a cos b + cos a sin b sin (a-b) = sin a cos b - cos a sin b product sin^2 a cos^2 b - cos^2 a sin^2 b now divide by cos^2 a cos^2 b sin^2 a/cos^2a - sin^2 b/ cos^2 b = tan^2 a - tan^2 b
*Friday, November 26, 2010 at 4:35pm by Damon*

**trig 30**

csc=1/sin cot=cos/sin so, csc^2-1/cotcsc = ((1/sin^2)-1)sin/(cos/sin) = sin((sin/sin^2)-sin)/(cos) = (sin^2/sin^2-sin^2)/cos = (1-sin^2)/cos = cos^2/cos = cos A ((sin^2/sin^2)-sin/(sin^2 cos)
*Wednesday, March 10, 2010 at 5:19pm by FredR*

**Trig**

1. sin x (cot x + tan x) = sin x (cos x/sin x + sin x/cos x) = cos x + (sin^2/cos x) = [cos^2 x + sin^2 x]/cos x = 1/cos x = sec x You try the other one
*Thursday, September 20, 2007 at 12:47am by drwls*

**Trig**

I think you mean sin x/(cos x + 1) + (cos x-1)/sin x = 0 multiply both sides by sin x (cos x+1) sin^2x + (cos x-1)(cos x+1)= 0 sin^2 x + cos^2 x - 1 = 0 but we know sin^2 x+cos^2 x = 1 1 - 1 = 0
*Saturday, March 3, 2012 at 6:09pm by Damon*

**Precal**

I do not understand how to do this problem ((sin^3 A + cos^3 A)/(sin A + cos A) ) = 1 - sin A cos A note that all the trig terms are closed right after there A's example sin A cos A = sin (A) cos (A) I wrote it out like this 0 = - sin^6 A - cos^6 A + 2sin^3 A cos^3 A - 2sin^3 ...
*Thursday, January 14, 2010 at 1:43pm by Joe*

**Pre-Calc/Math**

I think the parentheses are in the wrong place. It should probably read: (1-cos2x)/tan x = sin2x Again, split everything into sin and cos, and don't forget the identities: cos(2x)=cos²(x)-sin²(x) sin(2x)=2sin(x)cos(x) (1-cos(2x))/tan(x) =(1-(cos²(x)-sin²(x...
*Monday, May 14, 2012 at 7:15pm by MathMate*

**precalc**

((cosxcotx)/(1-sinx))-1=cscx (cos x*cos x/sin x) /(1-sin x) = 1 + 1/sin x (cos^2 x)/[sin x *(1-sin x)] = (sin x+1)/sin x cos^2 x/(1-sin x) = (1 + sin x) cos^2 x = (1-sin x)(1+sin x) cos^2 x = 1 - sin^2 x sure does
*Tuesday, April 1, 2008 at 4:23pm by Damon*

**trigonometry**

sec x = 1 / cos x sec x - cos x = 1 / cos x - cos x = 1 / cos x - cos ^ 2 x / cos x = ( 1 - cos ^ 2 x ) / cos x ________________________________________ Remark : 1 - cos ^ 2 x = sin ^ 2 x _______________________________________ 1 / cos x - cos ^ 2 x / cos x = ( 1 - cos ^ 2 x...
*Tuesday, December 4, 2012 at 12:19am by Bosnian*

**Trigonometry**

Identities: tan x = sin x / cos x sec x = 1 / cos x If your problem is this: (tanx/secx) + 1 Then: [(sin x / cos x) / (1 / cos x)] + 1 = sin x + 1 If your problem is this: tan x / (sec x + 1) Then: (sin x / cos x ) / [(1 / cos x) + 1] = (sin x / cos x) / [(1 / cos x) + (cos x...
*Wednesday, November 20, 2013 at 4:17pm by MathGuru*

**ALGEBRA**

Some identities: cos(a-b) = cos(a)cos(b) + sin(a)sin(b) sin(a-b) = sin(a)cos(b) - cos(a)sin(b) cot(a) = 1/tan(a) Therefore (substituting x for theta): cos(x-pi/2) = cos(x)cos(pi/2) + sin(x)sin(pi/2) ---------------------- sin(x-pi/2) = sin(x)cos(pi/2) - cos(x)sin(pi/2) cos(pi/...
*Tuesday, March 5, 2013 at 5:00pm by MathGuru*

**maths**

let A = x/2 , then we are proving tan^2 A = (1-cos(2A))/(1 + cos(2A)) LS = sin^2 A / cos^2 A RS = (1 - (cos^2 A - sin^2 A) )/(1 + cos^2 A - sin^2 A) = (1-cos^2 A + sin^2 A) / (1-sin^2 A + cos^2 A) = (sin^2 A + sin^2 A)/(cos^2 A + cos^2 A) = 2sin^2 A / (2cos^2 A) = sin^2 A/cos^...
*Friday, March 23, 2012 at 5:40am by Reiny*

**math trig**

cos/(1+sin) + (1+sin)/cos cos^2/[ cos(1+sin) ] +(1+sin)^2 /[cos(1+sin)] (cos^2 + 1 + 2 sin + sin^2)/[cos(1+sin)] but cos^2 + sin^2 = 1 ( 2 + 2 sin) /[cos(1+sin)] 2 (1+sin)/[cos(1+sin)] 2/cos which is 2 sec
*Wednesday, March 19, 2014 at 7:00pm by Damon*

**Trig**

left (sin/cos + cos/sin)^2 sin^2/cos^2 + 2 + cos^2/sin^2 [sin^4 +2sin^2 cos^2+cos^4 ]/cos^2 sin^2 (sin^2+cos^2)^2/cos^2sin^2 1^2/sin^2cos^2 1/sin^2 cos^2 right 1/cos^2 + 1/sin^2 sin ^2/cos^2sin^2 + cos^2/cos^2 sin^2 1/cos^2 sin^2
*Sunday, January 8, 2012 at 7:38pm by Damon*

**maths**

From the familiar sin(A+B)=sin(A)cos(B)+cos(A)sin(B), and sin(A-B)=sin(A)cos(B)-cos(A)sin(B) Add both equations to get sin(A+B)+sin(A-B)=2sin(A)cos(B) .....(3) But 2x=3x-x 4x=3x+x, so substitute A=3x, B=x into equation (3) and simplify to get your answer.
*Monday, December 24, 2012 at 8:00am by MathMate*

**calculus**

difference = y = cos ( t + pi/4) - cos t for part a take dy/dt and set to 0 dy/dt = -sin (t+pi/4) + sin t 0 = sin t - sin (t+pi/4) sin t = sin (t+pi/4) sin (a+b) =sin a cos b + cos a sin b so sin t = sin t cos pi/4 + cos t sin pi/4 sin t = sin t /sqrt 2 + cos t/sqrt 2 sqrt 2 ...
*Saturday, January 3, 2009 at 4:30pm by Damon*

**Trig**

Now we are cooking working on the left 1/cos t - cos t 1/cos t -cos^2 t/cos t (1 - cos^2 t)/cos t sin^2 t/cos t (sin t/cos t) sin t tan t sin t which is on the right
*Saturday, December 17, 2011 at 4:32pm by Damon*

**Trigonometry - simpler way**

using the half-angle formulas, you can show that 1 - cosA + cosB - cosA cosB = 2[sin^2 A/2 + cos^2 B/2 - cos^2 (A+B)/2] Then, with a little more fiddling, that can be changed to sin A/2 cos B/2 sin(A+B)/2 and the denominator comes out to cos A?2 sin B/2 sin(A+B)/2 divide those...
*Wednesday, January 4, 2012 at 9:31am by Steve*

**Math**

Solve this equation algebraically: (1-sin x)/cos x = cos x/(1+sin x) --- I know the answer is an identity, and when graphed, it looks like cot x. I just don't know how to get there. I tried multiplying each side by its conjugate, but I still feel stuck. This is what I have so ...
*Sunday, October 18, 2009 at 1:57pm by Momo*

**calc**

My answer, "|" = integral symbol |sin^4 x dx = -1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 | sin x dx ) -1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x + 1/2 (-cos x)) + C -1/4 sin^3 x cos x + 3/4 (-1/2 sin x cos x - 1/2 cos x) + C -1/4 sin^3 x cos x - 3/8 sin x cos x - 3/8 ...
*Sunday, February 20, 2011 at 7:56pm by Helper*

**Math**

Work on the left hand side: (1 + sinØ + cosØ)/(1 - sinØ + cosØ) multiply both the numerator and denominator by (1-sinφ+cosφ) and simplify: ((1+cosφ)^2-sin&hi;^2)/(1-sinφ+cosφ)^2 =2cosφ(1+cosφ)/[2(1-sinφ)(1+cosφ)] =2cosφ/[2(1-sinφ)] =...
*Tuesday, May 31, 2011 at 1:14pm by MathMate*

**trignonmetry**

PLEASE tell your classmates to use parentheses to clarify the problems! I think you mean (tan*cos^2 + sin^2)/sin = (sin*cos + sin^2)/sin = sin(cos+sin)/sin = cos+sin I think you mean (1+tan)/(1-tan) = (1+tan)^2/(1-tan^2) = (1+2tan+tan^2)/(1-tan^2) = (sec^2+2tan)/(1-tan^2) I ...
*Sunday, December 16, 2012 at 1:27pm by Steve*

**Check a few more CALC questions, please?**

#1. Since d/dx tan x = sec^2 x, you have 10x sec^2(5x^2) #2. ok #3. ok #4. y' = x^2-1 so, 0 #5. y = cos(1+sin) y' = [(-sin)(1+sin)-cos(cos)]/(1+sin)^2 = (-sin-sin^2-cos^2)/(1+sin)^2 = (-sin-1)/(1+sin)^2 = -1/(1+sin) #6. I. No idea what ^3(sqrt x-1) means II. y(1) = 1 or 2, so ...
*Thursday, December 19, 2013 at 11:58pm by Steve*

**precal- PLEASE HELP!**

suspect you might mean sec x / (sec x -tan x) = sec^2 x + sec x + tan x work on the left 1/cos x / [ 1/cos x -sin x/cos x ] = 1/[1 - sin x] = [ 1 + sin x ]/ [1-sin^2 x] = [1+sin x ]/cos^2 x = work on the right = 1/cos^2 x + 1/cos x + sin x/cos x) = (1/cos^2 x) [1 + cos x + sin...
*Sunday, March 25, 2012 at 9:21pm by Damon*

**trig**

Given: sin(x+y)=sin(x)cos(y) + cos(x)sin(y) For the case y=x: sin(x+x) = sin(x)cos(x) + cos(x)sin(x) =sin(2x) = 2sin(x)cos(x) or, just to make it easier 2sin(x)cos(x) = sin(2x) so, what does 4sin(x)cos(x) = ?
*Friday, June 12, 2009 at 1:00am by Quidditch*

**math**

how would you prove that sin^2(a)-cos^2(b)= sin^2(b)-cos^2(a). i'm not completely sure that this is right but i used the difference of two squares on it to get (sin(a)+cos(b))(sin(a)-cos(b)) then after that i am stuck. please help
*Monday, April 6, 2009 at 3:46pm by mike*

**advanced functions**

Rewrite the right side, using cos (x + y) = cos x cos y - sin x sin y cos (x - y) = cos x cos y + sin x sin y You should be familiar with those two identities.
*Sunday, November 8, 2009 at 10:29am by drwls*

**Precalculus**

That is simply the trig identity sin^2 x + cos^2 x = 1 You can show it easily with a right triangle with 90 deg at C where a^2 + b^2 = c^2 sin A = a/c cos A = b/c so sin^2 A = a^2/c^2 cos^2 A = b^2/c^2 and we know c^2 sin^2 A + c^2 cos^2 A = c^2 so sin^2 A + cos^2 A = 1
*Saturday, March 24, 2012 at 12:43pm by Damon*

**trig**

left side (sin/cos + 1/cos)^2 (1/cos^2)(sin^2 + 2 sin + 1) right side 2/cos^2 + 2 sin/cos^2 - cos^2/cos^2) (1/cos^2)( 2 + 2 sin -(1-sin^2) (1/cos^2)(2 + 2 sin -1 + sin^2) (1/cos^2)( sin^2 + 2 sin + 1) remarkable
*Monday, March 26, 2012 at 11:20am by Damon*

**calculus**

1/3 A(theta)= (0.5*(r^2)*theta)-(r*sin theta)= 0.5(theta-sin theta) B(theta)=(0.5*(r^2))*(0.5*r*cos theta* r*sin theta)=0.5sin theta(1- cos theta) so the limit as theta approaches zero from the positive side of a(theta) over b (theta) is equal to: [0.5(theta-sin theta)]/[0.5(...
*Sunday, November 8, 2009 at 5:02pm by Sam*

**Maths - due 2moro as well please help**

L.S. = cos^2 x/sin^2 x - cos^2 x = (cos^2 x - (sin^2 x)(cos^2 x))/sin^2 x = cos^2 x(1 - sin^2 x)/sin^2 x = cos^2 x(cos^2 x)/sin^2 x = (cos^2 x)(cot^2 x) = R.S.
*Saturday, July 5, 2008 at 3:19am by Reiny*

**Mathematics - Trigonometric Identities - Reiny**

Ok, let's start from the beginning sin^2x - sin^4x = cos^2x - cos^4x LS = (sinx - sin^2 x)(sinx + sin^2 x) then you had: (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) I will put in the in-between step =(sinx - (1 -cos^2x)) (sinx + 1 - cos^2x) = (sinx - 1 + cos^2x) (sinx + 1 - cos^2x...
*Saturday, November 10, 2007 at 6:50am by Reiny*

**Math**

In the future, please provide more information, it's nearly imposible to figure out what you're doing. Are you simplifying? It's difficult to tell. You may find it advantageous to know that: Sin^2+cos^2=1 thus 1-cos^2=sin^2 1-sin^2=cos^2 and tan= Sin/cos
*Wednesday, April 7, 2010 at 12:22pm by Houdini*

**Trig**

Use Lemma sin m + sin n = 2sin(m+n/2)cos(m-n/2) cos m + cos n = 2cos(m+n/2)cos(m-n/2) We get (sin m +sin n)/cos m + cos n) = sin(m+n/2)/cos(m+n/2) = tan m+n/2)
*Saturday, October 24, 2009 at 3:56pm by Arya*

**Math **

the original problem was: (sin x + cos x)^2 + (sin x - cos x)^2 = 2 steps too please I got 1 for (sin x + cos x)^2 but then what does (sin x - cos x)^2 become since it's minus?
*Saturday, March 19, 2011 at 10:00am by Amy*

**Trig**

Given: cos u = 3/5; 0 < u < pi/2 cos v = 5/13; 3pi/2 < v < 2pi Find: sin (v + u) cos (v - u) tan (v + u) First compute or list the cosine and sine of both u and v. Then use the combination rules sin (v + u) = sin u cos v + cos v sin u. cos (v - u) = cos u cos v + ...
*Friday, December 29, 2006 at 4:24pm by Nan*

**TRIG**

If sin s = -5/13, and it is in the third quadrant, cos s = -12/13 s = 202.620 degrees If sin t is -3/5, and t is in the fourth quadrant, cos t = 4/5. t = 323.130 degrees tan (s-t) = sin (s-t)/cos(s-t) = [sin s cos t - sin t cos s]/ [cos s cos t + sin s sin t] Crunch the ...
*Thursday, March 5, 2009 at 8:09pm by drwls*

**Math(Please check)**

Use the fundamental identities to simplify the expression. tan^2 Q / sec^2 Q sin^2/cos^2 / 1/cos^2 = sin^2 / cos^2 times cos^2 / 1 = The cos^2 cancels out so sin^2 is left. Is this correct?
*Friday, March 5, 2010 at 2:01pm by Hannah*

**math**

first change everything to sines and cosines 2/(sin^2 x) = 3(cos^2 x)/(sin^2 x) - 1 multiply by sin^2 x 2 = 3 cos^2 x - sin^2 x but sin^2 x - 1 - cos^2 x so 2 = 3 cos^2 x - 1 + cos^2 x 3 = 4cos^2 x cos^2 x = 3/4 cos x = ±√3/2 Do you recognize the 1,√3,2 right-...
*Tuesday, October 16, 2007 at 11:10pm by Reiny*

**Calculus I**

not quite if f = sin^2(u) f' = 2sin(u) cos(u) u' since u = e^v u' = e^v v' f = sin^2(e^sin^2 x) f' = 2 sin(e^sin^2 x)cos(e^sin^2 x) * e^(sin^2 x) * 2 sinx cos x = sin2x sin(2e^sin^2 x) e^(sin^2 x)
*Wednesday, March 21, 2012 at 1:41pm by Steve*

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