Question

Given that sin(2θ)=2/3, the value of

sin^6θ+cos^6θ

can be written as a/b with a and b coprime positive integers. Find a+b.

Answer 1

sin(2θ)=23


:O

Answer 2

are u sure sin(2 theeta) is 23 ?

Answer 3

no, actually i have now edited theques

Answer 4

\[\sin(2\theta)=\frac{2}{3}\]
\[\sqrt{1-\cos^{2}2\theta }=\frac{2}{3}\]
\[1-\cos^{2}2\theta=\frac{4}{9}\]
\[\cos^{2}2\theta=\frac{5}{9}\]
\[\cos2\theta=\frac{\sqrt{5}}{3}\]
\[1-2\sin^{2}\theta=\frac{\sqrt{5}}{3}\]
\[2\sin^{2}\theta=1-\frac{\sqrt{5}}{3}\]
\[\sin^{2}\theta=\frac{3-\sqrt{5}}{6}\]


\[\sin^{6}\theta +\cos^{6}\theta = \sin^{6}\theta + (\sqrt{1-\sin^{2}\theta})^{6}\]
\[\sin^{6}\theta+(1-\sin^{2})^{3}\]
\[\sin^{2}\theta+1-\sin^{6}\theta+3(1-\sin^{2}\theta)(-\sin^{2}\theta)\]
\[1-3\sin^{2}\theta+3\sin^{4}\theta\]

Substituting,
\[1-3(\frac{ 3-\sqrt{5} }{ 6 })+3(\frac{3-\sqrt{5}}{6})^{2}\]
\[1-(\frac{9-3\sqrt{5}}{6})+3(\frac{9+5-6\sqrt{5}}{36})\]
\[\frac{6-9+3\sqrt{5}+7-3\sqrt{5}}{6}\]
\[\frac{4}{6}\]
\[\frac{2}{3}\]

so, a+b= 2+3 =5