If the identity cos^4θ=acos4θ+bcos2θ+c holds for some a,b,c∈ℚ, then a,b,c=

Correct answer: (B) 1/8,1/2,3/8. Given cos^4θ=acos4θ+bcos2θ+c L.H.S.=cos^4θ=12+cos2θ2^2 =1/4+1/2cos2θ+1/4cos^22θ =1/4+1/2cos2θ+1/41/2+(cos4θ)/2 =1/8cos4θ+1/2cos2θ+3/8

Q49 / 160 · AP EAMCET 2022 (04 Jul Shift 1) · Mathematics

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

A
$(\frac{1}{8}, \frac{3}{8}, \frac{1}{2})$
Correct answer
B
$(\frac{1}{8}, \frac{1}{2}, \frac{3}{8})$
C
$(\frac{1}{2}, \frac{1}{8}, \frac{3}{8})$
D
$(\frac{1}{2}, \frac{3}{8}, \frac{1}{8})$

Solution

Given $\cos^{4} θ = a \cos 4 θ + b \cos 2 θ + c$
L.H.S. $= \cos^{4} θ = {(\frac{1}{2} + \frac{\cos 2 θ}{2})}^{2}$
$= \frac{1}{4} + \frac{1}{2} \cos 2 θ + \frac{1}{4} \cos^{2} 2 θ$
$= \frac{1}{4} + \frac{1}{2} \cos 2 θ + \frac{1}{4} (\frac{1}{2} + \frac{\cos 4 θ}{2})$
$= \frac{1}{8} \cos 4 θ + \frac{1}{2} \cos 2 θ + \frac{3}{8}$

Hence, the correct answer is option: 'B'.

If the identity cos4θ=acos4θ+bcos2θ+c holds for some a,b,c∈ℚ, then a,b,c=

Solution

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