Question

The expression $\dfrac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$ is equal to
$\eqalign{
  & 1)\cos 2x \cr
  & 2)2\cos x \cr
  & 3){\cos ^2}x \cr
  & 4)1 + \cos x \cr} $

Answer 1

Hint: The given question has only $\cos $ function in both the numerator and the denominator. So, we just need to group the terms and make sure that they cancel out each other. Since the options are also in terms of $\cos $, we can take a hint that we need to substitute any values for any of those functions.

Complete step by step solution:
The given expression is as follows,
$\dfrac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$
There are no terms common in the denominator, so we keep it as it is and group the terms in the numerator.
First, lets separate $6\cos 4x$ and $15\cos 2x$
$ = \dfrac{{\cos 6x + \cos 4x + 5\cos 4x + 5\cos 2x + 10\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$
Now, taking $5$and $10$common, we get
$ = \dfrac{{\cos 6x + \cos 4x + 5\left( {\cos 4x + \cos 2x} \right) + 10(\cos 2x + \cos 0)}}{{\cos 5x + 5\cos 3x + 10\cos x}}$
$1$ is written as $\cos 0$
Now, by applying the formula, we can simplify as
$ = \dfrac{{2\cos (5x)\cos (x) + 5(2\cos 3x\cos x) + 10(2\cos x\cos x)}}{{\cos 5x + 5\cos 3x + 10\cos x}}$
$2\cos x$ is common in all the terms. Therefore, taking $2\cos x$ out, we will have,
$ = \dfrac{{2\cos x\left( {\cos 5x + 5\cos 3x + 10\cos x} \right)}}{{\left( {\cos 5x + 5\cos 3x + 10\cos x} \right)}}$
Since the two expressions are similar, they will cancel each other out.
Therefore, the final answer is,
$\dfrac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}} = 2\cos x$
Hence, option (2) is the correct answer.

Additional Information:
The above used formula is called Compound Angle Formula. Compound angles are those that are formed by adding or subtracting two or more angles. There are compound angles formulas for sine and cosine functions. It is used when the trigonometric functions are to be added or subtracted.

Note: The above expression contains only positive functions and only cos functions. So, only the $\cos \left( {\alpha + \beta } \right)$ compound angle formula can be used. In order to use this formula, regroup the terms so that they have the coefficient. There are two similar options, be careful while choosing the right one.