Question

Prove :
$cos5\theta =16co{s}^5\theta -20co{s}^3\theta +5cos\theta$

Answer 1

By De-moivre's Theorem we know that
(cos $\theta$ + i sin $\theta {)}^5$ = cos 5$\theta$ + i sin 5$\theta$
L.H.S. on expansion by binomial theorem is
$co{s}^5\theta +5co{s}^4\theta \left(isin\theta \right)+10co{s}^3\theta (isin\theta {)}^2+10co{s}^2\theta (isin\theta {)}^3+5cos\theta (isin\theta {)}^4+(isin\theta {)}^5$
Now equate real and imaginary parts and change
$si{n}^2\theta to1-co{s}^2\theta andco{s}^2\theta to1-si{n}^2\theta$
depending on the answer.
$\therefore$ cos 5$\theta$ = cos $\theta$ (16 $co{s}^4\theta$ - 20 $co{s}^2\theta$ + 5)
sin = 5$\theta$ = sin $\theta$ (16 $si{n}^4\theta$ - 20 $si{n}^2\theta$ + 5)
Deduction : If $\theta$ = ${36}^{\circ }$, then $5\theta$ = ${180}^{\circ }$
$\therefore$ sin 5$\theta$ = 0
Also sin ${36}^{\circ }$ < sin ${45}^{\circ }$ or $si{n}^2{36}^{\circ }<\frac{1}{2}$
Now from (2), we get
0 = s(16 $s^4$ - 20 $s^2$ + 5), s = sin ${36}^{\circ }$ $\ne$ 0
$\therefore s^2=\frac{20\pm \sqrt{400-320}}{32}=\frac{10-2\sqrt{5}}{16}(\because s^2<\frac{1}{2})$
$\therefore s=\frac{\sqrt{10-2\sqrt{5}}}{5}$