Thursday

April 17, 2014

April 17, 2014

Number of results: 17,121

**Pre-Calc (Prove the identity)**

Recall that formula for sum in tangent: tan (a +/- b) = [tan(a) +/- tan(b)] / [1 -/+ tan(a)tan(b)] Applying this, tan^2(x/2 + π/4) = { [tan(x/2) + tan(π/4)] / [1 - tan(x/2)tan(π/4)] }^2 Note that tan(π/4) = 1, so = [(tan(x/2) + 1) / (1 - tan(x/2))]^2 = (tan...
*Sunday, November 24, 2013 at 4:29pm by Jai*

**Trig check my work please?**

Carry, where are you getting these, even though I enjoy doing them, they are getting to me, lol recall tan(A+B) = (tanA + tanB)/(1 - tanAtanB) so LS = (tanx + tan30)/1-tanxtan30)*(tan30-tanx)/(1+tanxtan30) = (tan^2 30 - tan^2 x)/(1 - (tan^x)(tan^2 30)) now tan^2 30º = 1/3 so ...
*Monday, February 23, 2009 at 8:45pm by Reiny*

**Trigonometry**

if you don't understand it at all, you seriously need to review the most basic trig identities: sin^2 + cos^2 = 1 tan = sin/cos cot = 1/tan With that in mind, and suppressing all the x's for ease of reading, (a) sin*tan = sin/cot = sin*cot/cot^2 = sin(cos/sin) / cot^2 = cos/...
*Sunday, November 24, 2013 at 4:10pm by Steve*

**Math**

Don't post a new question in the thread of a question. Tutors will see a reply and ignore it Start a new entry. if tanx = -7/24 and x is in IV then sin x = -7/25 and cosx = 24/25 cot x/2 = cos x/2 / sin x/2 recall cos 2A = 2cos^2 A - 1 or 1 - 2sin^2 A cos x = 2cos^ x/2 - 1 24/...
*Wednesday, January 30, 2013 at 6:10pm by Reiny*

**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*

**Trig**

Here's the whole deal: Let f(x) = (cos(x)+cos(2x)+...cos(6x)) / (sin(x)+...+sin(6x)) Multiply the denominator by 2sin(x/2) to get six products of 2sin(x/2)*sin(nx) where n=1 to 6. Using the sum and product formula, you should get: denominator = cos(x/2)-cos(3x/2)+ cos(3x/2)-...
*Tuesday, November 16, 2010 at 11:59am by MathMate*

**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*

**Derivatives**

H(x) = sin2x cos2x = 1/2 sin4x That help? Using the product rule, it's a bit more work, but you can get the same answer: dH/dx = 2cos2x cos2x - 2sin2x sin2x = 2(cos^2 2x - sin^2 2x) = 2cos4x Don't forget your trig just because you're taking calculus! #2 df/dx = (u'v - uv')/v^2...
*Thursday, October 3, 2013 at 3:52am by Steve*

**math**

I have a strong feeling you meant: (1-tan^2 Ø)/(1 + tan^2Ø) + 2tan^2 Ø = (1 - sin^2 Ø/cos^2 Ø)/sec^2 Ø + 2sin^2 Ø/cos^2 Ø = (1 - sin^2 Ø/cos^2 Ø)(cos^2 Ø) + 2sin^2 Ø/cos^2 Ø = cos^2 Ø - sin^2 Ø + 2sin^2 Ø/cos^2 Ø this is in terms of sines and cosines we could write it as cos (...
*Monday, February 17, 2014 at 11:29pm by Reiny*

**verifying identities**

how do you veriy tan x + Cot y -------------- = tan y + cot x 1-tan x cot y sin^4x + cos^4x = 1- 2cos^2x + 2cos^4
*Tuesday, July 22, 2008 at 11:09am by Dizzytrig*

**Pre-Calculus**

I don't understand,please be clear! Prove that each equation is an identity. I tried to do the problems, but I am stuck. 1. cos^4 t-sin^4 t=1-2sin^2 t 2. 1/cos s= csc^2 s - csc s cot s 3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x 4. sin^3 z cos^2 z= sin^3 z - sin^5 z
*Thursday, October 31, 2013 at 7:01pm by Anonymous*

**trig**

I would say they are the definitions of tan, cot, sec and csc in terms of sin and cos, tan = sin/cos cot = cos/sin csc = 1/sin sec = 1/cos sin^2 + cos^2 = 1 the law of sines the law of cosines sin (a + b) = sin a cos b + cos b sin a cos (a + b) = cos a cos b - sina sin b tan (...
*Monday, April 7, 2008 at 12:20am by drwls*

**Pre-Calculus**

Prove that each equation is an identity. I tried to do the problems, but I am stuck. 1. cos^4 t-sin^4 t=1-2sin^2 t 2. 1/cos s= csc^2 s - csc s cot s 3. (cos x/ sec x -1)- (cos x/ tan^2x)=cot^2 x 4. sin^3 z cos^2 z= sin^3 z - sin^5 z
*Thursday, October 31, 2013 at 10:03am by Anonymous*

**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*

**precalculus**

For each of the following determine whether or not it is an identity and prove your result. a. cos(x)sec(x)-sin^2(x)=cos^2(x) b. tan(x+(pi/4))= (tan(x)+1)/(1-tan(x)) c. (cos(x+y))/(cos(x-y))= (1-tan(x)tan(y))/(1+tan(x)tan(y)) d. (tan(x)+sin(x))/(1+cos(x))=tan(x) e. (sin(x-y...
*Sunday, April 14, 2013 at 3:27pm by anonymous*

**trig**

3sin^2(x)+sin(x)cos(x)=2sin^2(x)+2cos^2(x) sin^2(x)+sin(x)cos(x)-2cos^2(x)=0 Divide by cos^2(x) tan^2(x)+tan(x)-2=0 (tan(x)+2)(tan(x)-1)=0 tan(x)=-2 => x=-Arctan(2)+pi*n tan(x)=1 => x=pi/4+pi*n
*Saturday, September 10, 2011 at 4:56pm by Mgraph*

**Trigonometry**

Even though you are given two possible answers, you are expected to explain how your choice was made. Whenever you post an expression involving a division, you will need to put parentheses around the numerator and around the denominator to ensure an unambiguous definition of ...
*Saturday, January 2, 2010 at 7:27pm by MathMate*

**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*

**Pre Calc**

cotx = cos x/sin x tan x = sin x/ cos x cot x + tan x = cos x/sin x + sin x/ cos x so cot x + sin x = cos^2 x/sin x cos x + sin^2 x/sin x cos x cot x + tan x = (cos^2 x+sin^2x)/sin x cos x but cos^2 x + sin^2 x = 1 you take it from there.
*Tuesday, July 5, 2011 at 4:49pm by Damon*

**trig**

I do not understand these problems. :S I'd really appreciate the help. Use trigonometric identities to transform the left side of the equation into the right side. cot O sin O = cos O sin^2 O - cos^2O = 2sin^2 O -1 (tan O + cot O)/tan O = csc^2 O
*Monday, April 27, 2009 at 9:11pm by Matt*

**math**

We have to integrate sqrt[1+y'2] from 0 to pi/4 1+y'^2 = 1+tan^2(x) = 1/cos^2(x) So, the arc length is: Integral from 0 to pi/4 of 1/cos(x) = Integral from pi/4 to p1/2 of 1/sin(x) We can write: 1/sin(x) = 1/[2sin(1/2 x) cos(1/2 x)] = [sin^2(1/2x) + cos^2(1/2x)] /[2sin(1/2 x) ...
*Monday, December 3, 2007 at 4:25pm by Count Iblis*

**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*

**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**

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*

**Pre-Calculus**

So we must transform tan(A) + cot(A) into csc(A)*sec(A). tan(A) + cot(A) Note that cot(A) is equal to 1/tan(A). Substituting, = tan(A) + 1/tan(A) = (tan^2 (A) + 1) / tan(A) Recall the pythagorean identity: tan^2 (A) + 1 = sec^2 (A). Also tan(A) = sin(A)/cos(A) and sec(A) = 1/...
*Sunday, October 27, 2013 at 3:28pm by Jai*

**Trig**

using the identity cos 2A = cos^2 A - sin^2 A = 1 - 2sin^2 x = 2cos^2 x -1 LS = cos 4x + cos 2x = cos^2 2x - sin^2 2x + cos 2x = (1 - sin^2 (2x) ) - sin^2 (2x) + (1 - 2sin^2 x) = 1 - 2sin^2 (2x) + 1 - 2sin^2 x = 2 - 2sin^2 (2x) - 2sin^2 x = RS
*Sunday, February 17, 2013 at 11:33pm by Reiny*

**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*

**trigonometry**

tan(A-B) = (tanA - tanB)/(1+tanAtanB) but tanA = 2tanB so above = (2tanB - tanB)/(1 + 2tanBtanB) = tanB/(1+2tan^2B) = (sinB/cosB) / (1 + 2sin^2 B /cos^2 B) = sinB/cosB /[(cos^2 B + 2sin^2 B)/cos^2 B] = (sinB/cosB) * (cos^2 B)/(cos^2 B + 2sin^2 B) = sinBcosB /(cos^2 B + 2(1 - ...
*Tuesday, May 1, 2012 at 4:35pm by Reiny*

**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*

**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*

**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*

**Math - Solving Trig Equations**

What am I doing wrong? Equation: sin2x = 2cos2x Answers: 90 and 270 .... My Work: 2sin(x)cos(x) = 2cos(2x) sin(x) cos(x) = cos(2x) sin(x) cos(x) = 2cos^2(x) - 1 cos(x) (+/-)\sqrt{1 - cos^2(x)} = 2cos^2(x) - 1 cos^2(x)(1 - cos^2(x)) = 4cos^4(x) - 4cos^2(x) + 1 5cos^4(x) - 5cos^...
*Saturday, November 24, 2007 at 7:07pm by Anonymous*

**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*

**Trig**

1/cos + sin/cos = 1 1 + sin = cos 1 + 2sin + sin^2 = cos^2 = 1 - sin^2 2sin^2 + 2sin = 0 2sin(sin+1) = 0 sin=0 or sin = -1 x = 0, π, 3π/2 but, sec and tan are not defined at 3π/2, and π does not satisfy the original equation, so the solution is 0. so, x = ...
*Tuesday, May 15, 2012 at 4:20pm by Steve*

**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*

**Pre-Cal**

good observation recall that cos 2A = 2cos^2 A - 1 = 1 - 2sin^2 A so 1 - 2 cos^2 x = 2sin^2 x - 1 LS =sin x(1 - 2 cos^2 x + cos^4 x) = sinx(2sin^2 x - 1 + (cos^2x)(cos^2x) = sinx( 2 sin^2x - 1 + (1-sin^2x)(1 -sin^2x)) = sinx(2sin^2x - 1 + 1 - 2sin^2x + sin^4x) = sinx(sin^4x...
*Sunday, February 28, 2010 at 12:27pm by Reiny*

**verifying identities**

I get on the right: sin y / cos y + cos x / sin x (sin x sin y + cos x cos y) / sin x cos y sin x sin y/cos x + 1 cos y ---------------------------- sin x cos y/cos x tan x sin y + cos y --------------------- tan x cos y tan x sin y/sin y + cos y/sin y...
*Tuesday, July 22, 2008 at 11:09am by Damon*

**trigonometry**

cos(2x)-1=-2sin^2(x) sin(2x)=2sin(x)cos(x) The equation is -2sin(x)(sin(x)+cos(x))=0 sin(x)=0 or sin(x)+cos(x)=0 sin(x)=0==>x=pi,2pi,3pi,4pi sin(x)+cos(x)=0, sin(x)=-c0s(x) tan(x)=-1==>x=3pi/4,7pi/4,11pi/4,15pi/4, 19pi/4 (E)
*Thursday, May 12, 2011 at 7:45pm by Mgraph*

**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*

**math trigono**

1. cot = 1/tan 2. sec = 1/cos 7. tan,cot > 0 in QI,QIII 14. cos = sin/tan 19. 4^2 + (√17)^2 = (√33)^2, sin = tan*cos so, what are your answers?
*Wednesday, December 12, 2012 at 11:26am by Steve*

**Pre-calc/Calc**

Which expression is equivalent to sin(3x) + sin x? A. 2cos(2x)sin x B. 2sin(2x)sin x C. -2sin(2x)cos x D. 2sin(2x)cos x E. -2cos(2x)sin x
*Wednesday, April 17, 2013 at 11:12am by Chris Math Hw Help*

**Math**

7) since cos 2x = 2cos^2 x - 1, we have 6cos^2 x - 5cos x - 4 = 0 (2cos x + 1)(3cos x - 4) = 0 cos x = -1/2 --> x = 2pi/3 or 5pi/3 cos x = 4/3 --> no solution 8) (sin x - 2)(2sin x - 1) = 0 sin x = 2 --> no solution sin x = 1/2 --> x = pi/6 or 5pi/6 9) 3sec^2 x = 4...
*Thursday, March 22, 2012 at 11:26am by Steve*

**trig**

LS = (csc^2 Ø + cot^2 Ø)(csc^2 Ø - cot^2 Ø = (1 + cot^2 Ø + cot^2 Ø)(1+cot^2 Ø - cot^2 Ø = 1 + 2cot^2 Ø RS = (1 + cos^2 Ø)/sin^2 Ø = (sin^2 Ø + cos^2 Ø + cos^2 Ø)/sin^2 Ø = (sin^2 Ø + 2cos^2 Ø)/sin^2 Ø = 1 + 2 cot^2 Ø = LS
*Friday, January 24, 2014 at 8:46pm by Reiny*

**Trigonometry**

1.Solve tan^2x + tan x – 1 = 0 for the principal value(s) to two decimal places. 6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y 10.Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ 17.Prove that sin^2w-cos^2w/tan w sin w + cos w tan w = cos ...
*Tuesday, January 1, 2013 at 12:51pm by Alonso*

**maths**

since cot = cos/sin, sin cot = sin * cos/sin = cos expand the two tan expressions and recall that sec^2 = 1 + tan^2
*Monday, May 6, 2013 at 4:26pm by Steve*

**Trigonometry - Identities and proof**

Using the sum-to-product formulas, sinx-siny = 2cos((x+y)/2)sin((x-y)/2) cosx-cosy = -2sin((x+y)/2)sin((x-y)/2) now just divide to get cos((x+y)/2) / sin((x+y)/2) = cot((x+y)/2)
*Wednesday, November 13, 2013 at 11:23am 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*

**Math**

Which of the following are inverse functions? 1. Arcsin x and sin x 2. cos^-1 x and cos x 3. csc x and sin x 4. e^x and ln x 5. x^2 and +/- sqrt x 6. x^3 and cubic root of x 7. cot x and tan x 8. sin x and cos x 9. log x/3 and 3^x I believe the answers are 2, 4, 6, and 9, but ...
*Thursday, April 16, 2009 at 7:28pm by Anonymous*

**math (repost)**

Which of the following are inverse functions? 1. Arcsin x and sin x 2. cos^-1 x and cos x 3. csc x and sin x 4. e^x and ln x 5. x^2 and +/- sqrt x 6. x^3 and cubic root of x 7. cot x and tan x 8. sin x and cos x 9. log x/3 and 3^x I believe the answers are 2, 4, 6, and 9, but ...
*Thursday, April 16, 2009 at 9:02pm by Anonymous*

**Math**

tan (pi/4+x) = (tan pi/4 + tan x)/(1 -tan pi/4 tan x) tan (pi/4+x) = (tan pi/4 - tan x)/(1 +tan pi/4 tan x) note tan pi/4 = 1 add and put over common denominator [(1+tan x)(1+tan x)+(1-tan x)(1-tan x)]/(1 - tan^2 x) expand top [1+2 tanx +tan^2 x +1 - 2 tan x +tan^2 x]/(1 - tan...
*Sunday, January 4, 2009 at 2:48am by Damon*

**Calc**

Which functions of x and y in terms of time t can be derived from this rectangular equation? (x^2/4)+y^2=1 A. x = 2sin t, y = -cos t B. x = sin t, y = 5cos t C. x = sin t, y = 2cos t D. x = -sin t, y = 2cos t E. x = 2sin t, y = 5cos t
*Friday, April 26, 2013 at 11:27am by Chris*

**math**

I'll use x for theta, for my convenience. (cos^4x - sin^4x)/sin^2x cos^2x = (cos^2x - sin^2x)(cos^2x + sin^2x)/sin^2x cos^2x = (cos^2x - sin^2x)/sin^2x * (cos^2x + sin^2x)/cos^2x = (cot^2x - 1)(1+tan^2x) = cot^2x - 1 + cot^2xtan^2x - tan^2x = cot^2x - 1 + 1 - tan^2x = cot^2x...
*Wednesday, March 14, 2012 at 11:30pm by Steve*

**Calc**

Which expression is equivalent to sin(3x) + sin x? A. 2cos(2x)sin x B. 2sin(2x)sin x C. -2sin(2x)cos x D. 2sin(2x)cos x E. -2cos(2x)sin x
*Thursday, April 18, 2013 at 11:42am by Chris*

**Math**

Which expression is equivalent to sin(3x) + sin x? A) 2cos(2x)sin x B) 2sin(2x)sin x C) -2sin(2x)cos x D) 2sin(2x)cos x E) -2cos(2x)sin x
*Monday, March 24, 2014 at 10:57am by Anonymous*

**Pre-Cal**

Expand sin 2x using the identity: sin2x=2sin(x)cos(x) We'll get: 2cos(x)sin²(x)=cos(x) Transpose left-hand-side to the right: 2cos(x)((1/2)-sin²(x))=0 So cos(x)=0, or sin²(x)=1/2 Solve for x in (0,2π) as specified.
*Saturday, March 6, 2010 at 6:12pm by MathMate*

**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*

**trig**

since the tangent is negative, angle u is either in quadrant II or IV. I made a diagram of a right-angled triangle with a hypotenuse of √89 we know cos 2A = 2cos^2 A -1 or 1 - 2sin^2 A so cos u = 2cos^2 (u/2) - 1 -8/√89 + 1 = 2cos^2 (u/2) (√89 - 8)/(2√...
*Wednesday, March 3, 2010 at 3:00pm by Reiny*

**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*

**Pre Calc**

I don't think you really meant tan(4x) = 1 - sin^2(cos^2(x)), so do you mean tan 4x = 1 - 8 sin^2 x cos^2 x? Let's see. 8sin^2 x cos^2 x = 2*sin^2(2x)) tan(4x) =? 1 - 2sin^2(2x) Naah. Check your problem again. However, cos(4x) = 1 - 2sin^2(2x))
*Tuesday, November 6, 2012 at 2:04pm by Steve*

**trigonometry repost**

Reduce (csc^2 x - sec^2 X) to an expression containing only tan x. (is this correct?) csc x = 1/sin x sec x = 1/cos x tan x = 1/cot x sin^2 x + cos^2 x = 1 1 + cot^2 x = csc^2 x tan^2 x + 1 = sec^2 x csc^2 x - sec^2 x = 1 + cot^2 x - (1 + tan^2 x) = cot^2 x - tan^2 x = (1/tan^...
*Tuesday, February 1, 2011 at 8:54pm by Anonymous*

**Algebra 2 Honors**

LS = sin 2x(cot x + tan x) = 2 = 2sinxcosx(cosx/sinx + sinx/cosx) = 2cos^2x + 2sin^2x = 2(cos^2x + sin^2x) = 2(1) = 2 = RS
*Sunday, June 6, 2010 at 9:36pm by Reiny*

**Math (trigonometry)**

By inverting the fractions (a perfectly legal operation), the first equation can be converted to (sin x + cos x)/sin x = (1 + tan x)/x 1 + cot x = 1 + 1/tan x = 1 + cot x In the second problem, substitute 1 - cos^2 x for sin^2 x on the left side.
*Sunday, April 13, 2008 at 9:53pm by drwls*

**Math**

Notice that 67 degrees, 30' is half of 135 degrees, and 135 = 180-45 We also know that cos 2A = 2 cos^2 A - 1 giving us cos 135 = 2cos^2 67.5 - 1 So let's find cos 135 135 is in quadrant II, so cos 135 = -sin 45 = -1/√2 back to cos 135 = 2cos^2 67.5 - 1 -1/√2 = ...
*Monday, March 1, 2010 at 4:26pm by Reiny*

**trig**

Hint: When there is a mixture of tan, sin and cos, use identities to convert everything into sin and cos, and simplify. This usually works, especially if the right-hand side is in sin and cosine. For your information, tan(x)=sin(x)/cos(x) cot(x)=cos(x)/sin(x)
*Sunday, March 4, 2012 at 11:40am by MathMate*

**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*

**trig**

please help me with some questions I skipped on a review for our test coming up? solve 5-7 on the interval 0 greater than or equal to x less than or equal to 2pi. 5. sin x=sqrt3/2 6. cos x=-1/2 7. tan x=0 ---------------- 14. what is the exact value of tan[arcsin(.43...
*Wednesday, February 22, 2012 at 12:10pm by BreAnne*

**trigonometry hlp me**

6.Prove that tan y cos^2 y + sin^2y/sin y = cos y + sin y 10.Prove that 1+tanθ/1-tanθ = sec^2θ+2tanθ/1-tan^2θ 17.Prove that sin^2w-cos^2w/tan w sin w + cos w tan w = cos w-cot w cos w 23.Find a counterexample to shows that the equation sec a – cos a...
*Wednesday, January 2, 2013 at 1:15pm by eddy*

**Precalculus**

Not a surprise that you don't get it, the two expressions are not equivalent! Start from left: csc^4(x)-cot^4(x) Factor sin^4(x) as denominator: =(1-cos^4(x))/sin^4(x) Factor as difference of two squares = (1 + cos²(x)) (1 - cos²(x)) / sin^4(x) =(1 + cos²(x))sin...
*Thursday, December 9, 2010 at 6:45pm by MathMate*

**trig**

Unless you have found sin15 or cos15 before, and it has become part of your trig repertoire , you will have to do it twice since you must know that cos 30 = √3/2 and using cos 2A = 2cos^2 A - 1 , let 2A = 30 cos30 = 2cos^2 15 - 1 solving this for cos 15 I got √(&#...
*Sunday, October 17, 2010 at 2:05am by Reiny*

**trig**

Use double angle formula to reduce sin(2x) to 2sin(x)cos(x): 2sin(x)cos(x)cos(x)-sin(x)=0 sin(x)(2cos²(x)-1)=0 => sin(x)=0, or cos(x)=±1/√2 => Note: the interval was probably meant to be: [0,2π) x=0, x=π for sin(x)=0 or x=&pi/4, x=3π/4 for cos(...
*Sunday, August 14, 2011 at 3:07pm by MathMate*

**Trigonometry**

can't be factored because different functions are involved. As if you had y^2+x-1 = 0 But, using a well-known identity connecting sin and cos, 2cos^2 + sin - 1 = 0 2 - 2sin^2 + sin - 1 = 0 2sin^2 - sin - 1 = 0 (2sinx+1)(sinx-1) = 0
*Sunday, November 24, 2013 at 4:13pm by Steve*

**Math, derivatives**

f(x) = x² + 2Cos²x, find f ' (x) a) 2(x+cos x) b) x - sin x c) 2x + sin x d)2(x - sin2x) I got neither of these answers, since the 2nd part should be chain rule, right? f(x) = x² + 2Cos²x = x² + 2(Cos x)² then f '(x) = (2)(2)(cosx)(-sinx) My answer: f '(x) = 2x - 4cosxsinx If ...
*Monday, May 5, 2008 at 8:19pm by Terry*

**Trig**

sinӨ + sin2Ө = 0 recall that sin2Ө = 2sinӨcosӨ so sinӨ + sin2Ө = 0 sinӨ + 2sinӨcosӨ = 0 sinӨ(1+2cosӨ)=0 sinӨ = 0 or 2cosӨ = -1 sinӨ = 0 or or cosӨ = -1/2 Ө = 0,pi,2pi or &#...
*Wednesday, October 24, 2007 at 11:00am by Reiny*

**precalc**

cos 2x = 2cos^2 x - 1 so, 1/2 cos x (1+2cos^2 x - 1) = cos^3 x cos 4x = 1 - 2sin^2 2x = 1 - 8sin^2 x cos^2 x = 1 - 8sin^2 x (1 - sin^2 x) = 1 - 8sin^2 x + 8 sin^4 x cos 2x = 1 - 2sin^2 x 4cos 2x = 4 - 8sin^2 x 1/8(3-4cos2x+cos4x) = 1/8(3 - 4 + 8sin^2 x + 1 - 8sin^2 x + 8 sin^4...
*Thursday, August 16, 2012 at 11:24pm by Steve*

**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*

**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*

**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*

**Trigonometry**

Verify the identities. 1.) √1-COSθ/1+COSθ= 1+SINθ/SINθ 2.) SEC X SIN(π/2-X)= 1 3.) CSC X(CSC X-SIN X)+SIN X-COS X/SIN X + COT X= CSC²X 4.) CSC^4 X-2 CSC²X+1= COT^4 X 5.) CSC^4 θ-COT^4 θ= 2 CSC²θ-1 6.) TAN^5 X= TAN³X SEC²X-TAN³X 7...
*Sunday, February 17, 2013 at 1:42am by AwesomeGuy*

**trig**

RS = (3 - 4cos(2x) + cos^2(2x) - sin^2(2x))/8 = (3 - 4cos(2x) + cos^2(2x) - (1 - cos^2(2x)))/8 = (2cos^2(2x) - 4cos(2x) + 2)/8 = (2(cos^2(2x) - 2cos(2x) + 1))/8 = (2(cos(2x) - 1)^2)/8 = (2(1 - 2sin^2(x) - 1)^2)/8 = (2(-2sin^2(x))^2)/8 = (8sin^4(x))/8 = sin^4(x) = LS hope I ...
*Monday, February 23, 2009 at 8:15pm by Reiny*

**maths**

Expand both sides using sum of angles formula: cos(a+b)=cos(a)cos(b)-sin(a)sin(b) cos(a-b)=cos(a)cos(b)+sin(a)sin(b) Divide by cos(a) on both sides and solve for tan(a) to get: tan(a)=cot(b)*(1-m)/(1+m) Check me.
*Saturday, April 30, 2011 at 7:53am by MathMate*

**math**

cot 2x = 1/tan 2x = 1/[2tanx/(1-tan^2(x))] = (1-tan^2(x))/2tanx = [1 - 1/cot^2(x))/(2/cotx) = [(cot^2(x) - 1)/cot^2(x)]/[2/cotx] = (cot^2(x) - 1)/(2cotx) I don't know if that is the simplest way, I just sort of followed by nose. you can try sec 2x by noting sec 2x = 1/cos 2c ...
*Sunday, April 19, 2009 at 6:52pm by Reiny*

**MATH ANALYSIS**

It looks like you are working with the expansions of sin 2A and cos 2A, and are expected to know those formulas 1. is a direct result of the expansion of cos 2A = cos^2 A - sin^2 A so cos^2(x/2)-sin^2(x/2) = cosx 2. again, direct application 2cos^2(3A)-1 = cos 6A 3. sin 2x/1+...
*Tuesday, May 25, 2010 at 1:02am by Reiny*

**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*

**PreCalc**

Cleverly, you note that since tanθ = cotω, ω = π/2-θ Thus, θ-ω = θ-(π/2-θ) = π/2+2θ sin(θ-ω) = -cos2θ = 2sin^θ-1 tanθ = 9/5, so sinθ = 9/√106 sin(θ-ω) = 162/106 - 1 = 56/...
*Wednesday, May 8, 2013 at 12:57am by Steve*

**trigonometry**

1. LS = cos 3t = cos(2t + t) = cos2tcost - sin2tsint = (2cos^2 t - 1)(cost) - 2sintcostsint = 2cos^3 t - cost - 2sin^2 t cost = 2cos^3 t - cost - 2(1 - cos^2 t)cost = 2cos^3 t - cost - 2cost + 2cos^3 t = 4cos^3 t - 3cost = RS for #2, start with LS = sin(2x + 2x) = sin2xcos2x...
*Monday, August 27, 2012 at 2:22am by Reiny*

**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*

**Maths**

cos^2(360-A) = cos^2(A) tan^2(180+A) = tan^2(A) sec^2(180-A) = sec^2(A) sin^2(540+A) = sin^2(A) csc^2(-A) = csc^2(A) cot^2(90+A) = tan^2(A) so, using those simplifications, we have cos^2 * tan^2 * sec^2 = tan^2 sin^2 * csc^2 * tan^2 = tan^2 and the fraction is just 1 How far ...
*Thursday, May 2, 2013 at 11:18pm by Steve*

**college calculus**

Use the double angle identity to express cos(2x) in terms of sin(x) cos(2x) =cos^2(x)-sin^2(x) =1-2sin^2(x) sin(x)-2(1-sin^2(x))=0 2sin^2(x)+sin(x)-2=0 (2sin(x)-1)(sin(x)+1)=0 => sin(x)=1/2 or sin(x)=-1 Can you take it form here?
*Wednesday, February 29, 2012 at 11:48pm by MathMate*

**Math - Trig**

I'm trying to verify these trigonometric identities. 1. 1 / [sec(x) * tan(x)] = csc(x) - sin(x) 2. csc(x) - sin(x) = cos(x) * cot(x) 3. 1/tan(x) + 1/cot(x) = tan(x) + cot(x) 4. csc(-x)/sec(-x) = -cot(x)
*Monday, September 21, 2009 at 8:42pm by Liath*

**Pre-Cal**

cos e = sqrt (1 -sin^2e) = .995 tan e = sin e/cos e = .101 cot = 1/tan csc = 1/sin sec = 1/cos
*Thursday, December 15, 2011 at 10:50pm by Damon*

**trig**

Formulas to remember: sin(2a)=2sin(a)cos(a) cos(2a)=cos²(a)-sin²(a) From the second equation, replace a by a/2 and 2a by a to get cos(a)=cos²(a/2)-sin²(a/2) =2cos²(a/2)-1 ...(1) =1-2sin²(a/2) ...(2) From (1) we get cos²(a/2)=(1/2)(1+cos(a)) ...
*Wednesday, September 22, 2010 at 10:23pm by MathMate*

**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**

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*

**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*

**calculus**

There are standard formulae for integrals of rational functions of trigonometric formulae. In this case, you can simplify things as follows. Let's use the abbreviation: t = tan(x) s = sin(x) c = cos(x) We can write: (1-t)/(1+t) = (c-s)/(c+s) = (c - s)^2/(c^2 - s^2) = (c^2 + s^...
*Saturday, July 30, 2011 at 1:46pm by Count Iblis*

**trig/math**

if cosA = 12/13 in quadr IV, then sinA = -5/13 sin 2A = 2sinAcosA = 2(12/13)(-5/13) = -120/169 = cos 2A = cos^2 A - sin^2 A = 144/169 - 25/169 = 119/169 we know cos2A = 1 - 2sin^2 A or cos A = 1 - 2sin^2 (A/2) 12/13 = 1 - 2sin^2 (A/2) 2sin^2 (A/2) = 1 - 12/13 = 1/13 sin^2 (A/2...
*Friday, November 11, 2011 at 11:32am by Reiny*

**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*

**calculus**

Yo can find a reduction formula for the integral of tan^n(x) as follows. We have that tan^n(x) = sin^n(x)/cos^n(x) = sin^(n-2)(x)/cos^n(x) sin^2(x) = sin^(n-2)(x)/cos^n(x) [1-cos^2(x)] = sin^(n-2)(x)/cos^n(x) - sin^(n-2)(x)/cos^(n-2)(x) = sin^(n-2)(x)/cos^(n-2)(x) 1/cos^2(x...
*Thursday, June 3, 2010 at 11:22pm by Count Iblis*

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