The function g(x)=12x^2-sinx is the

Prove the following:[1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)]
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2. asked by Anonymous
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Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm(sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should
1. 3 answers
2. asked by Anonymous
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determine the absolute extreme values of the function f(x)=sinx-cosx+6 on the interval 0<=x<=2pi.This is what i did: 1.) i found
1. 3 answers
2. asked by Sara
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Simplify sin x cos^2x-sinxHere's my book's explanation which I don't totally follow sin x cos^2x-sinx=sinx(cos^2x-1)
1. 1 answer
2. asked by Tara
3. 1,050 views
Could someone check my reasoning? thanxFind the derivative of the function. sin(sin[sinx]) I need to use the chain rule to
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2. asked by XCS
3. 748 views
Hello! Can someone please check and see if I did this right? Thanks! :)Directions: Find the exact solutions of the equation in
1. 1 answer
2. asked by Maggie
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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^2x-sin^2x So it became
1. 1 answer
2. asked by Martha
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f(x)= (1-sinx) interval (0≤x≤2π) find where the function is increasing decreasing..i know that f`(x)=1-sinx it is equal to
1. 1 answer
2. asked by help
3. 563 views
tanx+secx=2cosx(sinx/cosx)+ (1/cosx)=2cosx (sinx+1)/cosx =2cosx multiplying both sides by cosx sinx + 1 =2cos^2x sinx+1 =
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2. asked by shan
3. 976 views
the problem is2cos^2x + sinx-1=0 the directions are to "use an identity to solve each equation on the interval [0,2pi). This is
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2. asked by alex
3. 860 views