Question

The expression
$
\dfrac{{cos6x + 6cos4x + 15cos2x + 10}}{{cos5x + 5cos3x + 10cosx}} \\
  \\
 $ is equal to
A.$cos2x$
B.$2cosx$
C.$co{s^2}x$
D.$1 + cosx$

Answer 1

Hint: 1.In the given question, we should first rearrange the question in such a manner that trigonometric identities can be applied to reduce the expression. We should use identities related to addition and subtraction.
2.Such questions are tricky and therefore students must be able to identify the terms associated with the variables in the questions.
3.Students must be able to comprehend these variables and establish a relationship between them to solve the question.
4.StudentsShould know the use of trigonometric expressions involved.
Here, we are making use of trigonometric identities.
$(\sin A + \operatorname{Sin} B),(\sin A - \sin B),(\cos A - \cos B)or(\cos A + \cos B)$
\[{\text{i}}{\text{.e cosA + cosB = 2cos(}}\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2})\]

Complete step-by-step answer:
Let us see what question demands,
$\dfrac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 3\cos 3x + 10\cos x}}$this is the left hand side of our equation.
To, solve this let us take the following approach
We should first split the term to take common to apply trigonometric identities of subtraction and addition such as
$(\sin A + \operatorname{Sin} B),(\sin A - \sin B),(\cos A - \cos B)or(\cos A + \cos B)$
Now
$\dfrac{{\cos 6x + (\cos 4x + 5\cos 4x) + (3\cos 2x + 10\cos 2x) + 10}}{{\cos 5x + 5\cos 3x + 10\cos x}}$
Now re-group therm to apply identities of addition and subtraction
$\dfrac{{(\cos 6x + \cos 4x) + (5\cos 4x + 5\cos 2x) + (10\cos 2x + 10\cos 0)}}{{\cos 5x + 5\cos 3x + 10\cos x}}$
Since
$
  [10 = 10 \times 1 = 10\cos 0] \\
  \dfrac{{(\cos 6x + \cos 4x) + 5(\cos 4x + \cos 2x) + 10(\cos 2x + \cos 0)}}{{\cos 5x + 5\cos 3x + 10\cos x}} \\
 $
 Let us apply trigonometry identity on each group in numerator
\[
  {\text{i}}{\text{.e cosA + cosB = 2cos(}}\dfrac{{A + B}}{2})\cos (\dfrac{{A - B}}{2}) \\
  \therefore {\text{ }}\dfrac{{2\cos (\dfrac{{6x + 4x}}{2}){\text{ }}\cos (\dfrac{{6x - 4x}}{2}) + 5(2{\text{ }}\cos (\dfrac{{4x + 2x}}{2}){\text{ }}\cos (\dfrac{{4x - 2x}}{2})) + 10(2\cos (\dfrac{{2x + 0}}{2}){\text{ }}\cos (\dfrac{{2x - 0}}{2}))}}{{\cos 5x + 5\cos 3x + 10\cos x}} \\
  = \,\,\dfrac{{2\cos 5x \times \cos x\,\, + \,\,5 \times (2\cos 3x.\cos x)\, + \,10 \times (2\cos x\,\cos x)}}{{\cos 5x + 5\cos 3x + 10\cos x}} \\
 \]Now take common terms in numerator and cancel it out with the denominator
$
 \Rightarrow \dfrac{{2\cos x[\{\cos 5x + 5\cos 3x + 10\cos x}]}{{\cos 5x + 5\cos 3x + 10\cos x}} \\
   = \,\,2\cos x \\
 $
$ \Rightarrow \dfrac{{\cos 6x + 6\cos 4x + 15\cos 2x + 10}}{{\cos 5x + 3\cos 3x + 10\cos x}} = 2\cos x$

So, the correct answer is “Option B”.

Note: 1.In such questions, we must not jump to simplify both numerator and denominator together. We should express which is more complicated or which can be reduced to another term present in the question.
2.In these types of examples the chances are that students can make mistakes in calculations so it is advised that they go step by step and not hurry considering the signs of the equations.
3. Students are advised to revise the trigonometric identities and operations associated with them.