If the identity cos^4 theta = a cos 4theta + | vedclass

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(B) Given $\cos^4 	heta = a \cos 4	heta + b \cos 2	heta + c$. Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we have: $\cos^4 	he

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$\left(\frac{1}{8}, \frac{1}{2}, \frac{3}{8}\right)$

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(B) Given $\cos^4 	heta = a \cos 4	heta + b \cos 2	heta + c$. Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we have: $\cos^4 	heta = \left(\frac{1 + \cos 2	heta}{2}ight)^2 = \frac{1}{4} (1 + 2 \cos 2	heta + \cos^2 2	heta)$. Using $\cos^2 2	heta = \frac{1 + \cos 4	heta}{2}$,we get: $\cos^4 	heta = \frac{1}{4} + \frac{1}{2} \cos 2	heta + \frac{1}{4} \left(\frac{1 + \cos 4	heta}{2}ight)$. $= \frac{1}{4} + \frac{1}{2} \cos 2	heta + \frac{1}{8} + \frac{1}{8} \cos 4	heta$. $= \frac{1}{8} \cos 4	heta + \frac{1}{2} \cos 2	heta + \frac{3}{8}$. Comparing this with $a \cos 4	heta + b \cos 2	heta + c$,we get $a = \frac{1}{8}$,$b = \frac{1}{2}$,and $c = \frac{3}{8}$. Thus,$(a, b, c) = \left(\frac{1}{8}, \frac{1}{2}, \frac{3}{8}ight)$.

If the identity $\cos^4 \theta = a \cos 4\theta + b \cos 2\theta + c$ holds for some $a, b, c \in \mathbb{Q}$,then $(a, b, c) =$

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