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Q. If $\alpha \cos ^2 3 \theta+\beta \cos ^4 \theta=16 \cos ^6 \theta+9 \cos ^2 \theta$ is an identity then -

Trigonometric Functions

Solution:

LHS $=\alpha\left(4 \cos ^3 \theta-3 \cos \theta\right)^2+\beta \cos ^4 \theta$
$=16 \alpha \cos ^6 \theta-24 \alpha \cos ^4 \theta+9 \alpha \cos ^2 \theta+\beta \cos ^4 \theta$
$=16 \alpha \cos ^6 \theta+(\beta-24 \alpha) \cos ^4 \theta+9 \alpha \cos ^2 \theta$
RHS $=16 \cos ^6 \theta+9 \cos ^2 \theta$
comparing LHS and RHS, we get $\alpha=1, \beta=24$