1
$\begingroup$

$$\frac{ \cos 2x}{1+\sin 2x}=\tan\left(\frac{\pi}{4}-x\right)$$

Hi. I'm confused how to prove this trig identity for the left side. If someone could help, then that would be great! I tried using $\sin(2x)=2\sin x\cos x$.

$\endgroup$
  • 1
    $\begingroup$ You don't "solve" an identity, you prove it. Start with the RHS. Use the formula for $\tan(x-y)$. Then, write everything in terms of sine and cosine and multiply both numerator and denominator by $(\cos x+\sin x)$. Recall the formulas for $\cos(2x)$ and $\sin(2x)$ and that $\sin^2 x+\cos^2 x=1$ and you're done. $\endgroup$ – learner Nov 28 '15 at 20:17
1
$\begingroup$

HINT: we have by the Addition formulas $$\frac{\cos(2x)}{1+\sin(2x)}=\frac{\cos(x)^2-\sin(x)^2}{1+2\sin(x)\cos(x)}=\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}$$

$\endgroup$
  • $\begingroup$ Thank you! How did you get from 1+2sin(x)cos(x) to cos(x) +sin(x) in the denominator? $\endgroup$ – melanie Nov 28 '15 at 20:42
  • $\begingroup$ Replace $1$ with $cos^2(x)+sin^2(x)$ from the Pythagorean theorem, so you now have $cos^2(x) + 2sin(x)cos(x) + sin^2(x)$ in the denominator. This looks like a quadratic that is a perfect square, or a binomial squared, that factors to $(cos(x) + sin(x))^2$. One of those factors then cancels with the numerator that factors to $(cos(x)+sin(x))(cos(x)-sin(x))$. $\endgroup$ – cr3 Nov 28 '15 at 21:54
1
$\begingroup$

L.H.S.=$\frac{\cos 2x}{1+\sin 2x}=\frac{\cos ^2 x-\sin ^2 x}{(\cos x+\sin x)^2}=\frac{\cos x-\sin x}{(\cos x+\sin x)}=\frac{1-\tan x}{1+ \tan x}=\tan(\frac{\pi}{4}-x)$=R.H.S.

$\endgroup$
1
$\begingroup$

Hints:

$$\cos 2x=\frac{1-\tan^2 x}{1+\tan^2 x}, \quad \sin 2x=\frac{2\tan x}{1+\tan^2 x},\quad\tan\Bigl(\frac\pi4-x\Bigr)=\frac{1-\tan x}{1+\tan x}.$$

$\endgroup$

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.