1
$\begingroup$

I have to show that the left equation simplifies to $\tan\theta$:

Show that: $$\frac{1-\cos2 \theta}{\sin2 \theta} = \tan \theta$$

I do have prior knowledge that: $$\tan \theta = \frac{\cos \theta}{\sin \theta}$$

But I'm stuck from this point, I have tried a few rules, but none have seemed to work so far.

$\endgroup$
2
$\begingroup$

Using the double angle formulae:

$$\cos(2\theta)=1-2\sin^2\theta$$ $$\sin(2\theta)=2\sin\theta\cos\theta$$

$$\frac{1-\cos(2\theta)}{\sin(2\theta)}=\frac{1-1+2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{2\sin^2\theta}{2\sin\theta\cos\theta}=\frac{\sin\theta}{\cos\theta}=\tan\theta$$

$\endgroup$
1
$\begingroup$

you need the double angle formulae $ cos2\theta=2cos^2\theta-1$ and $ sin2\theta=2sin\theta cos\theta $

$\endgroup$
  • $\begingroup$ you can prove those from sin(a+b) cos(a+b) expressions $\endgroup$ – VigneshM Nov 11 '14 at 7:35
1
$\begingroup$

use $cos2\theta$=$1-2{sin^2{\theta}}$ and $sin2\theta$=$2{sin\theta}{cos\theta}$ and you will get your required answer.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.