Saturday
April 19, 2014

Search: Math sin/cos

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

Pages: 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | 11 | 12 | 13 | 14 | 15 | Next>>

Search
Members