# Value of $\cos^2\alpha-\sin^2\alpha$

My problem is from Israel Gelfand's Trigonometry textbook.

Page 48. Exercise 8: b) If $\tan\alpha=r$, write an expression in terms of $r$ that represents the value of $\cos^2\alpha-\sin^2\alpha$.

The attempt at a solution:

Well I solved a) which was similar, question was: a) If $\tan\alpha=2/5$, find the numerical value of $\cos^2\alpha-\sin^2\alpha$. Solution was this, since $\tan\alpha=\frac{\sin\alpha}{\cos\alpha}=\frac{2}{5}$, then $\cos^2\alpha-\sin^2\alpha=5^2-2^2=21$. b) part is more general version of a) I guess, but I can't seem to find identity that would allow me to solve it. I would appreciate some hints, thank you in advance.

• $\frac ab=\frac cd$ does not mean that $a=c$ and $b=d$ !! – Yves Daoust Jul 21 '14 at 6:27
• So my solution to a) is not correct? – George Apriashvili Jul 21 '14 at 6:31
• @GeorgeDirac No, because $\cos\alpha$ can not exceed 1. – Bernhard Jul 21 '14 at 6:39
• yeah that's true of course, looks like I have to redo it, thank you @Yves Daoust for noticing my mistake. – George Apriashvili Jul 21 '14 at 6:43
• Well, the numerical value would be $21/29=0.724$. – George Apriashvili Jul 21 '14 at 6:49

Two Hints:

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

• Thank you very much, I understand now. Answer would be $\frac{1-r^2}{1+r^2}$. – George Apriashvili Jul 21 '14 at 6:29
• @GeorgeDirac - yes it would :) – nbubis Jul 21 '14 at 6:34

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

• Using Mathjax on this site is not only preferred, but demanded. – 5xum Jul 21 '14 at 6:18
• Very elegant and economical! – Yves Daoust Jul 21 '14 at 9:17

We have

$$\cos^2\alpha-\sin^2\alpha=2\cos^2\alpha-1$$ and by $$\cos^2\alpha=\frac1{1+\tan^2\alpha}$$ we find

$$\cos^2\alpha-\sin^2\alpha=\frac2{1+r^2}-1=\frac{1-r^2}{1+r^2}$$

We have: $$\sin \alpha=\pm\frac{\tan \alpha}{\sqrt{1 + \tan^2 \alpha}}\!$$ and $$\cos \alpha=\pm\frac{1}{\sqrt{1 + \tan^2 \alpha}}\!$$ then $$\sin^2 \alpha=\frac{\tan^2 \alpha}{1 + \tan^2 \alpha}=\frac{r^2}{1 + r^2}\!$$ and $$\cos^2 \alpha=\frac{1}{1 + \tan^2 \alpha}\!=\frac{1}{1 + r^2}$$

You start from $r=\frac{\sin\alpha}{\cos\alpha}$ and must evaluate $\cos^2\alpha-{\sin^2\alpha}$.

This should push you to set $c=\cos^2\alpha, s=\sin^2\alpha$, use $$c+s=1$$ and rewrite the first relation as $$r^2c-s=0.$$ Solve this linear system for $c$ and $s$: $$(1+r^2)\ c=1\\(1+r^2)\ s=r^2.$$ Then$$c-s=\frac{1-r^2}{1+r^2}.$$

($\sin\theta$/$\cos\theta$)=r
or,($\cos\theta$/$\sin\theta$)=1/r
or,($\cos^2\theta$ )/($\sin^2\theta$)=1/$r^2$[squaringbothsides]
or,($\cos^2\theta$+$\sin^2\theta$)/($\cos^2\theta$-$\sin^2\theta$)=(1 + $r^2$)/(1 - $r^2$)[ApplyingComponendoandDividendo]
or,1/($\cos^2\theta$- $\sin^2\theta$)=(1 + $r^2$)/(1 - $r^2$ )
or,($\cos^2\theta$- $\sin^2\theta$)=(1 - $r^2$)/(1 + $r^2$)

• Please use Mathjax, otherwise it's very hard to see the solution. – George Apriashvili Jul 21 '14 at 6:38
• For starters Here is the tutorial! – MonK Jul 21 '14 at 9:30

My Approach:

$\tan \alpha=r$

$\implies \tan^2 \alpha=r^2$

$\implies \large \frac{\sin^2 \alpha}{\cos^2 \alpha}=r^2$

$\implies \large \frac{\sin^2 \alpha+\cos^2 \alpha}{\sin^2 \alpha-\cos^2 \alpha}=\large \frac{r^2+1}{r^2-1}$

$\implies \large \frac{1}{\sin^2 \alpha - \cos ^2 \alpha}=\large \frac{r^2+1}{r^2-1}$

$\implies {\sin^2 \alpha - \cos ^2 \alpha}=\large \frac{r^2-1}{r^2+1}$

Multiply both sides by -1

$\implies {\cos ^2 \alpha-\sin^2 \alpha}=\large \frac{1-r^2}{1+r^2}$