# Problem on Trigonometry

If $\sin x + \cos x = a$ , then find:

(i) $\sin^6 x + \cos^6 x$

(ii) $\lvert \sin x – \cos x \rvert$

I tried using the identity $a^3 + b^3$ for (i) but that didn't work and I ended up with an expression that was of no help. I have absolutely no idea on how to go about the problem. Any kind of help would be appreciated.

Given $\displaystyle \sin x+\cos x = a\Rightarrow (\sin x+\cos x)^2 = a^2\Rightarrow \sin x \cos x = \frac{a^2-1}{2}$

Using $(\sin ^2 x+\cos^2 x)^3 = (\sin^6 x+\cos^6 x)+3(\sin x\cos x)^2(\sin^2 x+\cos^2 x)$

So $\displaystyle 1=\sin^6 x+\cos^6 x+3\left(\frac{a^2-1}{2}\right)$

$\displaystyle \bullet\; |\sin x-\cos x|^2 = (\sin x-\cos x)^2=1-2\sin x\cos x = 1-\frac{2(a^2-1)}{2}$

We have $$\sin^6 x +\cos^6 x =(\sin^2 x +\cos^2 x)(\sin^4 x + \cos^4 x -\sin^2 x \cos^2 x)$$ $$=(1)((\sin^2 x + \cos^2 x)^2-3\sin^2 x \cos^2 x)$$ $$=(1-3\sin^2 x\cos^2 x) \tag {1}$$

Now we know that $$(\sin x + \cos x)^2 =1 +2\sin x \cos x$$ Use this result to find the value of $(1)$.

For $(2)$, use the fact that $$(\sin x -\cos x)^2 =1-2\sin x\cos x$$ Hope it helps.

• I used the same approach for (i) but then couldn't figure out how to use the given information to find it's value. I had absolutely no idea about (ii). Well, thanks a lot. – SamInuyasha ANMF Jan 29 '17 at 7:38