Question

sin θ cos(900-θ) + cos θ sin (900-θ) =

Answer 1

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Answer 2

Answer:

Sin θ cos(90°-θ) + cos θ sin (90°-θ) = 1

Sin θ cos(900°-θ) + cos θ sin (900°-θ) = 0

Step-by-step explanation:

Sin θ cos(90°-θ) + cos θ sin (90°-θ)

Using formula Sin(A + B) = SinACosB + CosASinB

= Sin(θ + 90°-θ)

= Sin90°

= 1

Sin θ cos(90°-θ) + cos θ sin (90°-θ) = 1

or

Sin θ cos(90°-θ) + cos θ sin (90°-θ)

using cos(90°-θ) = Sinθ  & sin (90°-θ) = Cosθ

= SinθSinθ + cos θ cos θ

= Sin²θ + cos²θ

= 1

or if we consider it 900°

Sin θ cos(900°-θ) + cos θ sin (900°-θ)

900° = 2*360° + 180°

= Sin θ cos(180°-θ) + cos θ sin (180°-θ)

= Sin(θ + 180°-θ)

= Sin 180°

= 0

Sin θ cos(900°-θ) + cos θ sin (900°-θ) = 0