2
$\begingroup$

Knowing: $$ \tan x=2-\sqrt{3} $$

Obtain: $$\cos2x$$

I tried converting $\tan x$ into it's sinus and cosine form and trying to square both sides to try to get the form of:

$$\cos^2x-\sin^2x$$

But I can't really get to this form without having extra expressions of sine or cosine, any ideas how to start this properly?

Taken out of one of the entry tests to Maths in TAU.

Solution:

$$\cos2x=\cos^2x-\sin^2x=\frac{\cos^2x-\sin^2x}{\sin^2x+\cos^2x}:\frac{\cos^2x}{\cos^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$

$$ \frac{1-\tan^2x}{1+\tan^2x}=\frac{-6+4\sqrt{3}}{8-4\sqrt{3}}=\frac{-3+2\sqrt{3}}{4-2\sqrt{3}} $$

$\endgroup$
  • $\begingroup$ Hint form a right triangle assuming the adjacent side to be 1 $\endgroup$ – Karl Sep 9 '18 at 12:50
  • $\begingroup$ You sound as if you have done all you need if you have both $\sin x$ and $\cos x$. Why not edit your question with that working and the chances are you have just made a silly error that someone will spot for you $\endgroup$ – Mandelbrot Sep 9 '18 at 12:51
4
$\begingroup$

HINT

Recall that by half-angle identities

$$\cos 2x = \frac{1-\tan^2 x}{1+\tan^2 x}$$

$\endgroup$
  • $\begingroup$ Your ship travels directly, but mine has double routes. :-)) $\endgroup$ – mrs Sep 9 '18 at 12:53
  • $\begingroup$ @mrs Yes you gave the way to obtain that! It's nice :) $\endgroup$ – gimusi Sep 9 '18 at 12:53
5
$\begingroup$

One may use the following identities:

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

  2. $\cos(2x)=2\cos^2(x)-1$

$\endgroup$
3
$\begingroup$

HINT:

$$\tan^2x=\frac{\sin^2x}{\cos^2x}=(2-\sqrt3)^2=13-4\sqrt3$$ and $$\cos2x=\cos^2x-\sin^2x=\cos^2x(1-\tan^2x)$$ and use this with the identity $1+\tan^2x=\sec^2x$

$\endgroup$

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.