Please check proof of identity

**Furyan** · January 14th, 2012, 03:11 PM

Hello all

The question is show that:

$\dfrac{\sin 2\theta}{1+\cos 2\theta} \equiv \tan\theta$

I have

$LHS \equiv\dfrac{\sin\theta\cos\theta + \cos\theta\sin\theta}{1 + \cos^2\theta - \sin^2\theta}$

$\equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}$

$\equiv\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}$

$\equiv\dfrac{\sin\theta}{\cos\theta} \equiv \tan\theta$

Is that alright and is the a better way?

Thank you.

**e^(i*pi)** · January 14th, 2012, 03:39 PM

Originally Posted by Furyan

Hello all

The question is show that:

$\dfrac{\sin 2\theta}{1+\cos 2\theta} \equiv \tan\theta$

I have

$LHS \equiv\dfrac{\sin\theta\cos\theta + \cos\theta\sin\theta}{1 + \cos^2\theta - \sin^2\theta}$

$\equiv\dfrac{2\sin\theta\cos\theta}{1 + \cos^2\theta - (1 - cos^2\theta)}$

$\equiv\dfrac{2\sin\theta\cos\theta}{2\cos^2\theta}$

$\equiv\dfrac{\sin\theta}{\cos\theta} \equiv \tan\theta$

Is that alright and is the a better way?

Thank you.

That way is fine. I'd have omitted the third line of working since the reader should be able to establish that you converted the existing $1-\sin^2 \theta$ into $\cos^2 \theta$

Spoiler:

**Furyan** · January 14th, 2012, 03:49 PM

Originally Posted by e^(i*pi)

That way is fine. I'd have omitted the third line of working since the reader should be able to establish that you converted the existing $1-\sin^2 \theta$ into $\cos^2 \theta$

Spoiler:

Thank you very much indeed, I didn't see that. I keep missing simple things like that. Thank you for checking my work and pointing out that step.

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