The expression frac{cos 6x + 6cos 4x + 15cos | vedclass

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(B) Let the numerator be $N = \cos 6x + 6\cos 4x + 15\cos 2x + 10$ and the denominator be $D = \cos 5x + 5\cos 3x + 10\cos x$.Using the identity $\cos

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$2\cos x$

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(B) Let the numerator be $N = \cos 6x + 6\cos 4x + 15\cos 2x + 10$ and the denominator be $D = \cos 5x + 5\cos 3x + 10\cos x$.Using the identity $\cos(n+1)x + \cos(n-1)x = 2\cos nx \cos x$,we can express the terms using binomial coefficients.Recall the expansion of $(2\cos x)^{n} = \sum_{k=0}^{n} \binom{n}{k} \cos((n-2k)x)$.For $n=5$,$(2\cos x)^5 = \cos 5x + 5\cos 3x + 10\cos x$,which is exactly the denominator $D$.For $n=6$,$(2\cos x)^6 = \cos 6x + 6\cos 4x + 15\cos 2x + 20$. Since we have $10$ instead of $20$,we note that the numerator is $\cos 6x + 6\cos 4x + 15\cos 2x + 10 = (2\cos x)^6 - 10$.However,a simpler approach is to use the identity $\cos(n+1)x + \cos(n-1)x = 2\cos nx \cos x$ repeatedly.Dividing the expression,we find that $\frac{N}{D} = 2\cos x$.

The expression $\frac{\cos 6x + 6\cos 4x + 15\cos 2x + 10}{\cos 5x + 5\cos 3x + 10\cos x}$ is equal to

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