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Exercise 2.2 · Q31

Q.Find sin⁡θ\sin\theta such that 3cos⁡θ+4sin⁡θ=43\cos\theta + 4\sin\theta = 4.

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Step 1. 3cos⁡θ+4sin⁡θ=4⇒cos⁡θ=4−4sin⁡θ33\cos\theta+4\sin\theta=4 \Rightarrow \cos\theta=\dfrac{4-4\sin\theta}3.

Step 2. Substitute into cos⁡2θ=1−sin⁡2θ\cos^2\theta=1-\sin^2\theta: 16(1−sin⁡θ)29=1−sin⁡2θ=(1−sin⁡θ)(1+sin⁡θ)\dfrac{16(1-\sin\theta)^2}9 = 1-\sin^2\theta = (1-\sin\theta)(1+\sin\theta).

Step 3. For sin⁡θ≠1\sin\theta\ne1, divide both sides by (1−sin⁡θ)(1-\sin\theta): 16(1−sin⁡θ)9=1+sin⁡θ⇒16−16sin⁡θ=9+9sin⁡θ⇒7=25sin⁡θ⇒sin⁡θ=725\dfrac{16(1-\sin\theta)}9=1+\sin\theta \Rightarrow 16-16\sin\theta=9+9\sin\theta \Rightarrow 7=25\sin\theta \Rightarrow \sin\theta=\dfrac7{25}. …

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