Question

If sin2θ + sin2(null) =½ and cos2θ + cos2(null) =3/2 then cos²(θ-null)=

a)3/8
b)5/8
c) ¾
d)5/4
don't spam pls and pls answer by 2:00pm IST 09 Nov 21​

Answer 1

the answer is 2) 5/8

Solution:

Given,

sin 2θ + sin 2Φ = 1/2….(i)

cos 2θ + cos 2Φ = 3/2….(ii)

Squaring and adding (i) and (ii),

sin2(2θ) + sin2(2Φ) + 2 sin 2θ sin 2Φ + cos2(2θ) + cos2(2Φ) + 2 cos 2θ cos 2Φ = (1/4) + (9/4)

[sin2(2θ) + cos2(2θ)] + [sin2(2Φ) + cos2(2Φ)] + 2[cos 2θ cos 2Φ + sin 2θ sin 2Φ] = 10/4

Using the identities sin2A + cos2A = 1 and cos(A – B) = cos A cos B – sin A sin B,

1 + 1 + 2 cos(2θ – 2Φ) = 5/2

2[1 + cos 2(θ – Φ)] = 5/2

Using the formula 1 + cos 2A = 2 cos2A,

2 cos2(θ – Φ) = 5/4

cos2(θ – Φ) = 5/8

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