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Exercise 3.2 · Q38

Q.Prove the following: cos⁡θ+sin⁡(270°+θ)−sin⁡(270°−θ)+cos⁡(180°+θ)=0\cos\theta+\sin(270°+\theta)-\sin(270°-\theta)+\cos(180°+\theta)=0

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Step 1: sin⁡(270°+θ)=sin⁡270°cos⁡θ+cos⁡270°sin⁡θ=(−1)cos⁡θ+0=−cos⁡θ\sin(270°+\theta)=\sin270°\cos\theta+\cos270°\sin\theta=(-1)\cos\theta+0=-\cos\theta.

Step 2: sin⁡(270°−θ)=sin⁡270°cos⁡θ−cos⁡270°sin⁡θ=−cos⁡θ\sin(270°-\theta)=\sin270°\cos\theta-\cos270°\sin\theta=-\cos\theta.

Step 3: cos⁡(180°+θ)=−cos⁡θ\cos(180°+\theta)=-\cos\theta. …

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