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

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(B) Let the numerator be $N = \cos 6x + 6\cos 4x + 15\cos 2x + 10$. Using the identity $\cos nx = 2\cos((n-1)x)\cos x - \cos((n-2)x)$,or by expanding

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$2\cos x$

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(B) Let the numerator be $N = \cos 6x + 6\cos 4x + 15\cos 2x + 10$. Using the identity $\cos nx = 2\cos((n-1)x)\cos x - \cos((n-2)x)$,or by expanding using binomial coefficients,we observe that the numerator is the real part of $(e^{ix} + e^{-ix})^6 / 2^6 	imes 2^6$ or simply recognizing the expansion of $(2\cos x)^6$. Alternatively,using the identity $\cos(nx) = \sum_{k=0}^{n} \binom{n}{k} \cos((n-2k)x)$,we can see that the numerator is $(2\cos x)^6 / 2 = 32\cos^6 x$. However,a simpler way is to use the expansion of $(2\cos x)^n$. We know that $(2\cos x)^6 = 2\cos 6x + 12\cos 4x + 30\cos 2x + 20$. Dividing by $2$,we get $\cos 6x + 6\cos 4x + 15\cos 2x + 10 = \frac{(2\cos x)^6}{2} = 32\cos^6 x$. Similarly,for the denominator $D = \cos 5x + 5\cos 3x + 10\cos x$,we know $(2\cos x)^5 = 2\cos 5x + 10\cos 3x + 20\cos x$. Dividing by $2$,we get $\cos 5x + 5\cos 3x + 10\cos x = \frac{(2\cos x)^5}{2} = 16\cos^5 x$. Therefore,the expression is $\frac{32\cos^6 x}{16\cos^5 x} = 2\cos x$.

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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