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Question: If \(\sin\theta_{1} + \sin\theta_{2} + \sin\theta_{3} = 3,\) then \(\cos\theta_{1} + \cos\theta_{2} ...

If sin⁡θ1+sin⁡θ2+sin⁡θ3=3,\sin\theta_{1} + \sin\theta_{2} + \sin\theta_{3} = 3, then cos⁡θ1+cos⁡θ2+cos⁡θ3=\cos\theta_{1} + \cos\theta_{2} + \cos\theta_{3} =

A

3

B

2

C

1

D

0

Answer

0

Explanation

Solution

sin⁡θ1+sin⁡θ2+sin⁡θ3=3\sin\theta_{1} + \sin\theta_{2} + \sin\theta_{3} = 3

⇒sin⁡θ1=sin⁡θ2=sin⁡θ3=1\Rightarrow \sin\theta_{1} = \sin\theta_{2} = \sin\theta_{3} = 1, (∵−1≤sin⁡x≤1)(\because - 1 \leq \sin x \leq 1)

⇒θ1=θ2=θ3=π2⇒cos⁡θ1+cos⁡θ2+cos⁡θ3=0\Rightarrow \theta_{1} = \theta_{2} = \theta_{3} = \frac{\pi}{2} \Rightarrow \cos\theta_{1} + \cos\theta_{2} + \cos\theta_{3} = 0.