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The expression cos^2 q + cos^2 (a+q) - 2 cos a cosq cos (a+q) is independent of

(A) q
(b) a
(C) both a and q
(D)none of a and q
7 years ago

Fawz Naim
37 Points

cos^2 q+cos^2(a+q)-2cosa.cosq.cos(a+q)

cos^2 q+cos(a+q)[cos(a+q)-2cosa.cosq]

cos^2 q+cos(a+q)[cosa.cosq-sina.sinq-2cosa.cosq]

cos^2 q+cos(a+q)[-cosa.cosq-sina.sinq]

cos^2 q-cos(a+q)[cosa.cosq+sina.sinq]

cos^2 q-cos(a+q).[cos(a-q)]

cos^2 q-[2cos(a+q).cos(a-q)/2]

cos^2 q-[cos2a+cos2q/2]

2cos^2 q-cos2a-cos2q/2

2cos^2 q-(cos^2 a-sin^2 a)-(cos^2 q-sin^2 q)/2

2cos^2 q-cos^2 a+sin^2 a-cos^2q+sin^2 q/2

cos^2 q+sin^2 q-cos^2 a+sin^2 a/2

1-cos^2 a+sin^2 a/2

therefore it is independent of q.So option (A) is correct

7 years ago
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