Let the range of the function f(x)=6+16 x (( )/(3)-x) (( )/(3)+x) 3x 6x,x R be [ , ].

Question

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

Accepted Answer

Using $\cos x\cos\left(\dfrac\pi3-x\right)\cos\left(\dfrac\pi3+x\right) = \dfrac14\cos3x$:

$$f(x) = 6+16\cdot\frac14\cos3x\cdot\sin3x\cdot\cos6x = 6+4\cos3x\sin3x\cos6x = 6+2\sin6x\cos6x = 6+\sin12x$$

Since $\sin12x \in [-1,1]$, the range of $f$ is $[5,7]$, so $(\alpha,\beta) = (5,7)$.

$$\text{Distance} = \frac{|3(5)+4(7)+12|}{\sqrt{3^2+4^2}} = \frac{|15+28+12|}{5} = \frac{55}{5} = 11$$

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