
 0
 0
posted by Steve
Respond to this Question
Similar Questions

Calculus
Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help. Calculus  Steve, Tuesday, January 12, 2016 at 
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  
Integral
That's the same as the integral of sin^2 x dx. Use integration by parts. Let sin x = u and sin x dx = dv v = cos x du = cos x dx The integral is u v  integral of v du = sinx cosx + integral of cos^2 dx which can be rewritten 
Math (trigonometric identities)
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 
Trig
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 
Math
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 
Trig Help
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. 
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 
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 
Mathematics  Trigonometric Identities
Prove: sin^2x  sin^4x = cos^2x  cos^4x What I have, LS = (sinx  sin^2x) (sinx + sin^2x) = (sinx  1 cos^2x) (sinx + 1  cos^2x) = sin^2x + sinx  sinx  cos^2xsinx  cos^2xsinx  1  1 + cos^4x = sin^2x  2cos^2xsinx  2 +