# Thread: [SOLVED] Exercises frustration! (proving trig identities)

1. ## [SOLVED] Exercises frustration! (proving trig identities)

Hi again fellow math gurus (minus "fellow")!

I'm trying to prove that:

1) sin(pi/2 - x) = sin(pi/2 + x)

This is how I went about it...

RHS = sin(pi/2) cos x + cos(pi/2) sin x
= (1) cos x
= cos x

I know that doesn't exactly work out and I'm stumped hahaha!

2) This one is more recent:

(2 tan x)/(1+tan^2 x) = sin 2x

I'm not looking for the answer so much as maybe a hint at what identities to start with? I tried going about it couple different ways but this stuff is just greek to me hehe. I'm trying to get really good at it before my final exam in 3 weeks.

Thanks for the continued support!

2. Originally Posted by DannyMath
Hi again fellow math gurus (minus "fellow")!

I'm trying to prove that:

1) sin(pi/2 - x) = sin(pi/2 + x)

This is how I went about it...

RHS = sin(pi/2) cos x + cos(pi/2) sin x
= (1) cos x
= cos x

I know that doesn't exactly work out and I'm stumped hahaha!

Use in both sides the angles sum identities: $\displaystyle \sin(x\pm y)=\sin x\cos y \pm \sin y\cos x$

2) This one is more recent:

(2 tan x)/(1+tan^2 x) = sin 2x

$\displaystyle \frac{2\tan x}{1+\tan^2x}=\frac{2\sin x\slash \cos x}{1+\sin^2x\slash\cos^2x}=\frac{2\sin x\cos x}{\cos^2x+\sin^2x}$ , and remember the duplication angle identity $\displaystyle \sin 2x=2\sin x\cos x$ ( and, of course, the

trigonometric Pythagoras' Theorem).

Tonio

I'm not looking for the answer so much as maybe a hint at what identities to start with? I tried going about it couple different ways but this stuff is just greek to me hehe. I'm trying to get really good at it before my final exam in 3 weeks.

Thanks for the continued support!
.

3. Originally Posted by DannyMath
Hi again fellow math gurus (minus "fellow")!

I'm trying to prove that:

1) sin(pi/2 - x) = sin(pi/2 + x)

This is how I went about it...

RHS = sin(pi/2) cos x + cos(pi/2) sin x
= (1) cos x
= cos x

I know that doesn't exactly work out and I'm stumped hahaha!

2) This one is more recent:

(2 tan x)/(1+tan^2 x) = sin 2x

I'm not looking for the answer so much as maybe a hint at what identities to start with? I tried going about it couple different ways but this stuff is just greek to me hehe. I'm trying to get really good at it before my final exam in 3 weeks.

Thanks for the continued support!
For 2.

$\displaystyle \tan(x) = \frac{\sin(x)}{\cos(x)}$ and hence $\displaystyle \tan^2(x) = \frac{\sin^2(x)}{\cos^2(x)}$

So $\displaystyle \frac{2\tan(x)}{1 + \tan^2(x)} = \frac{2\tfrac{sin(x)}{cos(x)}}{1 + \tfrac{sin^2(x)}{cos^2(x)}}$.

Now note that...
$\displaystyle 1 + \frac{\sin^2(x)}{\cos^2(x)} = \frac{cos^2(x)}{cos^2(x)} + \frac{\sin^2(x)}{\cos^2(x)}$.
$\displaystyle = \frac{1}{\cos^2(x)}$.

So substituting this in gets you...

$\displaystyle \frac{2\tfrac{sin(x)}{cos(x)}}{1 + \tfrac{sin^2(x)}{cos^2(x)}} = 2\sin(x)\cos(x) = \sin(2x)$.

EDIT: Too slow...

4. Ok, I understand the second problem, thanks guys!

For the first one, if I enter the addition formula on both sides, I end up with:

LHS = sin (pi/2) cos x - cos (pi/2) sin x
RHS = sin (pi/2) cos x + cos (pi/2) sin x

But these aren't equivalent are they?

5. Originally Posted by DannyMath
Ok, I understand the second problem, thanks guys!

For the first one, if I enter the addition formula on both sides, I end up with:

LHS = sin (pi/2) cos x - cos (pi/2) sin x
RHS = sin (pi/2) cos x + cos (pi/2) sin x

But these aren't equivalent are they?
$\displaystyle \cos(\tfrac{\pi}{2}) = 0$

$\displaystyle \sin(\tfrac{\pi}{2}) = 1$

Substitute those into the equations and see what's left.

6. Originally Posted by DannyMath

But these aren't equivalent are they?
I'm funny! What an unfortunate way for me to finish my assignment, by forgetting to apply fundamental trig knowledge! :P

Thank you!