Q. 1 B4.3( 12 Votes )

Evaluate the following :

B. (sinθ + cosθ)2 + (sinθ - cosθ)2

Answer :

We have

(sin θ + cos θ)2 + (sin θ – cos θ)2 = ((sin θ + cos θ) + (sin θ – cos θ))2 – 2(sin θ + cos θ)(sin θ – cos θ) [, a2 + b2 = (a + b)2 – 2ab]


(sin θ + cos θ)2 + (sin θ – cos θ)2 = (sin θ + cos θ + sin θ – cos θ)2 – 2(sin2 θ – cos2 θ)


[, (a + b)(a – b) = a2 – b2]


(sin θ + cos θ)2 + (sin θ – cos θ)2 = (2 sin θ)2 – 2 sin2 θ + 2 cos2 θ


(sin θ + cos θ)2 + (sin θ – cos θ)2 = 4 sin2 θ – 2 sin2 θ + 2 cos2 θ


(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 sin2 θ + 2 cos2θ


(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 (sin2 θ + cos2 θ)


(sin θ + cos θ)2 + (sin θ – cos θ)2 = 2 [, sin2 θ + cos2 θ = 1]


Thus, (sin θ + cos θ)2 + (sin θ – cos θ)2 = 2.


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