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Question : 9 of 160
Marks:
+1,
-0
Solution:
cos4θ=a⋅cos‌4‌θ+b‌cos‌2‌θ+c ‌⇒‌‌cos4θ=a[2cos22θ−1]+b[2cos2θ−1]+c
‌⇒‌‌cos4θ=a⋅2cos22θ−a+2bcos2θ−b+c
‌⇒‌‌cos4θ=2a(cos‌2‌θ)2+2bcos2θ−a−b+c
‌⇒‌‌cos4θ=2a(2cos2θ−1)2+2bcos2θ−a−b+c
‌⇒‌‌cos4θ=2a(4cos4θ+1−4cos2θ)+2bcos2θ−a−b+c
⇒cos4θ=8acos4θ‌+2a−8acos2θ‌+2bcos2θ−a−b+c
⇒cos4θ=8acos4θ+cos2θ(−8a+2b)+a−b+c
On comparing the corresponding coefficients, we get
8a=1,−8a+2b=0 and
a−b+c=0 ⇒‌‌a=‌ ∴‌−8(‌)+2b=0 gives
b=‌ ‌ and ‌‌‌−‌+c=0 gives
c=‌
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