Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I need to prove this:

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

using only $\sin(a-b)$ and $\cos(a-b)$ formulas wich I already proved. I also proven this:

$$\cos^2(x) + \sin^2(x) = 1$$ $$\sin(-x) = -\sin(x)$$ $$\cos(2x) = \cos^2x -\sin^2(x)$$ $$\sin(2x) = 2\sin(x)\cos(x)$$ and the $\sin(a+b)$ and $\cos(a+b)$ formulas. Do I need something more so I can prove this identity? Could you guys give me some help?

share|improve this question
1  
I recommend setting $x=2y$ and substituting the variable $y$ in for $x$. –  NotNotLogical Apr 11 at 3:18

1 Answer 1

up vote 6 down vote accepted

$$\cos2x=\cos^2x-\sin^2x=\frac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$ dividing the numerator & the denominator by $\cos^2x$


Dividing the numerator & the denominator by $\cos^2x$ $$\cos2x=\cos^2x-\sin^2x=\frac{1-\tan^2x}{\sec^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$

share|improve this answer
    
why did you divide by $cos^2(x) + \sin^2(x)$? –  Marter Js Apr 11 at 3:20
1  
Because that’s equal to $1$. –  Lubin Apr 11 at 3:23

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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