If {cos ^6}α + {sin ^6}α + K,{sin ^2}2α = 1, then K =

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3.Trigonometrical Ratios, Functions and Identities

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If ${\cos ^6}\alpha + {\sin ^6}\alpha + K\,{\sin ^2}2\alpha = 1,$ then $K =$

A

$\frac{4}{3}$

B

$\frac{3}{4}$

C

$\frac{1}{2}$

D

$2$

Solution

(b) Since ${\cos ^6}\alpha + {\sin ^6}\alpha + K{\sin ^2}2\alpha = 1$

using formula ${a^3} + {b^3} = {(a + b)^3} – 3ab(a + b)$ and on solving,

we get the required result $i.e.$ $K = \frac{3}{4}$.

Standard 11

Mathematics

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