Proving/Solving trig equations

**live_laugh_luv27** · April 19th 2009, 05:05 PM

1) Solve
(sinx)/(cot^2x) - (sinx)/(cos^2x)

2) Prove
tan^2x - sin^2x = tan^2x*sin^2x

3) Prove
sin(x - π/2) = -cosx

Thanks for any help!

**skeeter** · April 19th 2009, 05:30 PM

Originally Posted by live_laugh_luv27

1) Solve
(sinx)/(cot^2x) - (sinx)/(cos^2x)

2) Prove
tan^2x - sin^2x = tan^2x*sin^2x

3) Prove
sin(x - π/2) = -cosx

Thanks for any help!

#1 can only be simplified ... it's not an equation.

$\sin{x}\left(\frac{1}{\cot^2{x}} - \frac{1}{\cos^2{x}}\right)$

$\sin{x}(\tan^2{x} - \sec^2{x})$

$\sin{x}[\tan^2{x} - (1 + \tan^2{x})] = -\sin{x}$

#2 change the right side to $\tan^2{x}(1 - \cos^2{x})$ ... work from there to get the left side.

#3 use the difference identity for sine on the left side ...

$\sin(a - b) = \sin(a)\cos(b) - \cos(a)\sin(b)<br />$

**live_laugh_luv27** · April 19th 2009, 05:36 PM

Thanks so much..you've been really helpful

**live_laugh_luv27** · April 19th 2009, 06:22 PM

sorry, i just have one question.
Could you please explain how you did #2? I can't get past the second step.
How did you get tan^2x(1-cos2x) to equal tan^2x - sin^2x? After I distribute the tan^2x, I'm not sure what to do.
Thanks!

**skeeter** · April 19th 2009, 06:31 PM

Originally Posted by live_laugh_luv27

sorry, i just have one question.
Could you please explain how you did #2? I can't get past the second step.
How did you get tan^2x(1-cos2x) to equal tan^2x - sin^2x? After I distribute the tan^2x, I'm not sure what to do.
Thanks!

change $tan^2{x}$ into $\frac{\sin^2{x}}{\cos^2{x}}$ , then distribute.

**live_laugh_luv27** · April 19th 2009, 06:38 PM

great, thanks!

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