Let the range of the function f(x) = 6 + 16 c | vedclass

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(A) Using the identity $\cos 	heta \cos(\frac{\pi}{3} - 	heta) \cos(\frac{\pi}{3} + 	heta) = \frac{1}{4} \cos 3	heta$,we have: $f(x) = 6 + 16 \lef

Vedclass · Accepted Answer

$11$

Vedclass · Answer

(A) Using the identity $\cos 	heta \cos(\frac{\pi}{3} - 	heta) \cos(\frac{\pi}{3} + 	heta) = \frac{1}{4} \cos 3	heta$,we have: $f(x) = 6 + 16 \left(\frac{1}{4} \cos 3xight) \sin 3x \cdot \cos 6x$ $f(x) = 6 + 4 \cos 3x \sin 3x \cos 6x$ Using $2 \sin 	heta \cos 	heta = \sin 2	heta$,we get $4 \cos 3x \sin 3x = 2 \sin 6x$: $f(x) = 6 + 2 \sin 6x \cos 6x$ Using $2 \sin 	heta \cos 	heta = \sin 2	heta$ again: $f(x) = 6 + \sin 12x$ Since $-1 \le \sin 12x \le 1$,the range of $f(x)$ is $[6-1, 6+1] = [5, 7]$. Thus,$\alpha = 5$ and $\beta = 7$. The point is $(5, 7)$. The distance from the point $(x_1, y_1)$ to the line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. $d = \frac{|3(5) + 4(7) + 12|}{\sqrt{3^2 + 4^2}} = \frac{|15 + 28 + 12|}{\sqrt{25}} = \frac{55}{5} = 11$.

Let the range of the function $f(x) = 6 + 16 \cos x \cdot \cos \left(\frac{\pi}{3} - x\right) \cdot \cos \left(\frac{\pi}{3} + x\right) \sin 3x \cdot \cos 6x$,where $x \in R$,be $[\alpha, \beta]$. Then the distance of the point $(\alpha, \beta)$ from the line $3x + 4y + 12 = 0$ is:

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