Trigonometry
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Mathematics  Trigonometric Identities  Reiny
Mathematics  Trigonometric Identities  Reiny, Friday, November 9, 2007 at 10:30pm (sinx  1 cos^2x) (sinx + 1  cos^2x) should have been (sinx  1 + cos^2x) (sinx + 1  cos^2x) and then the next line should be sin^2x + sinx 
asked by Anonymous on November 10, 2007 
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Hello! Can someone please check and see if I did this right? Thanks! :) Directions: Find the exact solutions of the equation in the interval [0,2pi] cos2x+sinx=0 My answer: cos2x+sinx=cos^2xsin^2x+sinx =1sin^2xsin^2x+sinx
asked by Maggie on February 2, 2015 
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I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx1/sinx+1 = cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x  5 = 4tan^2x  1 20. Cosx
asked by Kim on December 4, 2012 
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Simplify sin x cos^2xsinx Here's my book's explanation which I don't totally follow sin x cos^2xsinx=sinx(cos^2x1) =sinx(1cos^2x) =sinx(sin^2x) (Where does sine come from and what happend to cosine?) =sin^3x
asked by Tara on January 4, 2010 
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I need help solving for all solutions for this problem: cos 2x+ sin x= 0 I substituted cos 2x for cos^2xsin^2x So it became cos^2(x)sin^2(x) +sinx=0 Then i did 1sin^2(x)sin^2(x)+sinx=0 = 12sin^2(x)+sinx=0 = sinx(2sinx+1)=1
asked by Martha on February 26, 2017 
calculus
Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the
asked by XCS on May 7, 2009 
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1.)Find the exact solution algebriacally, if possible: (PLEASE SHOW ALL STEPS) sin 2x  sin x = 0 2.) GIVEN: sin u = 3/5, 0 < u < ï/2 Find the exact values of: sin 2u, cos 2u and tan 2u using the doubleangle formulas. 3.)Use the
asked by MAKITA on April 10, 2009 
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1)tan Q = 3/4 Find cosQ 3^2 + 4^2 = x^2 9+16 = sqrt 25 = 5 cos = ad/hy = 4/5 Am I correct? 2) Use the sum and difference identites sin[x + pi/4] + sin[xpi/4] = 1 sinx cospi/4 + cosxsin pi/4 + sinx cos pi/4  cosx sin pi/4 =
asked by Abbey on March 16, 2010 
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Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1cos^2x)]/[sinx+1] =??? This is where I'm stuck. Can someone help me.
asked by Anonymous on March 11, 2012 
Trigonometry.
( tanx/1cotx )+ (cotx/1tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated
asked by mo on April 18, 2007