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Exercise 11.5 · Q8

Q.sin⁡2x1+cos⁡x\dfrac{\sin^2 x}{1+\cos x}

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sin⁡2x=1−cos⁡2x\sin^2x=1-\cos^2x factors as a difference of squares that cancels the denominator exactly.

Step 1. Factor the numerator. sin⁡2x=1−cos⁡2x=(1−cos⁡x)(1+cos⁡x)\sin^2x=1-\cos^2x=(1-\cos x)(1+\cos x).

Step 2. Cancel the common factor. (1−cos⁡x)(1+cos⁡x)1+cos⁡x=1−cos⁡x\dfrac{(1-\cos x)(1+\cos x)}{1+\cos x}=1-\cos x. …

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