Question

Find the value of $p$ such that $\cos x\cos (240 - x)\cos (240 + x) \in [ - p,p]$
A.$\dfrac{1}{2}$
B.$\dfrac{1}{3}$
C.$\dfrac{1}{4}$
D.$\dfrac{3}{2}$

Answer 1

The given expression has trigonometric functions with similar angles, so we can modify it converting two or three functions into one function of double or triple angle. This will reduce the number of functions and thus make it much easier to solve.

Complete step-by-step answer:
The given expression is $\cos x\cos (240 - x)\cos (240 + x)$ .
Here the angle of cosine looks very similar to the standard formula of $2\cos C\cos D = \cos (C + D) + \cos (C - D)$ .
To convert the expression in to the standard form, we first have to multiply and divide the expression by $2$, which gives us $\dfrac{{\cos x}}{2}[2\cos (240 - x)\cos (240 + x)]$ .
So the value of $2\cos (240 - x)\cos (240 + x)$ becomes

$$\cos (240 - x + 240 - x) + \cos (240 + x - (240 - x)) \\\ = \cos (480) + \cos (2x) \\\$$

Now, the overall expression can be written as $\dfrac{{\cos x}}{2}[\cos (480) + \cos (2x)]$ .
We can simplify this further by substituting the value of $\cos (480) = \cos (360 + 120) = \cos (120) = - \dfrac{1}{2}$ .
The expression is now written as $\dfrac{{\cos x}}{2}[ - \dfrac{1}{2} + \cos (2x)]$ .
Now, as we know the value of $\cos (2x)$ is $2{\cos ^2}x - 1$, we can substitute this too in the expression.
This gives the expression as $\dfrac{{\cos x}}{2}[ - \dfrac{1}{2} + 2{\cos ^2}x - 1]$ which is same as $\dfrac{{\cos x}}{2}[ - \dfrac{3}{2} + 2{\cos ^2}x]$.
To simplify this further, we multiply the $\dfrac{{\cos x}}{2}$ inside the bracket to give $- \dfrac{{3\cos x}}{4} + {\cos ^3}x$.
This further simplification gives $\dfrac{{4{{\cos }^3}x - 3\cos x}}{4}$.
The expression obtained above, if you recall, has the formula for triple angle in cosine function in numerator.
As we know $\cos 3x = 4{\cos ^3}x - 3\cos x$, this can be substituted in the numerator to give, $\dfrac{{\cos 3x}}{4}$.
This is the most simplified form of the expression and from it we can easily determine its range.
As we know $\cos x$ as well $\cos 3x$ have a range from $- 1$ to $1$, so $\dfrac{{\cos x}}{4}$ as well as $\dfrac{{\cos 3x}}{4}$ will have a range from $- \dfrac{1}{4}$ to $\dfrac{1}{4}$.
Now as the range was $[ - p,p]$, so the value of $p$ will be $\dfrac{1}{4}$.
So, option (C) is correct.

Note: The formula for double angle of cosine has many variations, but we use the one which has cosine function, so that in the expression we have only one type of trigonometric function that is cosine. Selecting another variation of double angle of cosine will complicate the expression leading to mistakes.
Also, notice how in the final step, $4$ is in the denominator of cosine function and not in the denominator of its angle, that’s why the end values of range become one-fourth. If the $4$ would have been in the denominator of the angle of cosine, then there would have been no change in end values of range from the initial value.