Question

The value of trigonometric expression$\dfrac{{(\sin {\text{ }}4x{\text{ }} + \sin {\text{ }}3x{\text{ }} + {\text{ }}\sin 2x)}}{{(\cos {\text{ }}4x{\text{ }} + {\text{ }}\cos {\text{ }}3x{\text{ }} + \cos {\text{ }}2x)}}{\text{ }}$ is equal to-
a). $\cot 3x$
b). $\cos 3x$
c). $\tan 3x$
d). None of these

Answer 1

To solve this question first we assume the value of the given expression. For further solving use the formula of $\sin c + \sin d$ and $\cos c + \cos d$ because we see here $\sin 3x$ and $\cos 3x$ apply the formula on $\sin 4x + \sin 2x$ and $\cos 4x + \cos 2x$ because from her we get some common terms. After taking common terms out we get the same terms in numerator and the denominator now cancels that term and converting that term to another trigonometric function we get the final answer.
$\sin c + \sin d = 2\sin (\dfrac{{c + d}}{2})\cos (\dfrac{{c - d}}{2})$,
$\cos c + \cos d = 2\cos (\dfrac{{c + d}}{2})\cos (\dfrac{{c - d}}{2})$, and
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$

Complete step-by-step solution:
Let $t = \dfrac{{(\sin 4x{\text{ }} + \sin 3x{\text{ }} + {\text{ }}\sin 2x)}}{{(\cos 4x{\text{ }} + {\text{ }}\cos 3x{\text{ }} + \cos 2x)}}{\text{ }}$
For further solving we use the formula $\sin c + \sin d = 2\sin (\dfrac{{c + d}}{2})\cos (\dfrac{{c - d}}{2})$ and $\cos c + \cos d = 2\cos (\dfrac{{c + d}}{2})\cos (\dfrac{{c - d}}{2})$
We use these formulas in $\sin 4x + \sin 2x$ and $\cos 4x + \cos 2x$ because from here we get the terms of $\sin 3x$ and $\cos 3x$.
On using these formula
$t = \dfrac{{\sin 3x{\text{ }} + {\text{ }}2\sin \left( {\dfrac{{2x + 4x}}{2}} \right)\cos \left( {\dfrac{{4x - 2x}}{2}} \right)}}{{\cos 3x{\text{ }} + 2\cos \left( {\dfrac{{2x + 4x}}{2}} \right)\cos \left( {\dfrac{{4x - 2x}}{2}} \right)}}{\text{ }}$
On further solving
$t = \dfrac{{\sin {\text{ }}3x{\text{ }} + {\text{ }}2\sin \left( {3x} \right)\cos \left( x \right)}}{{\cos {\text{ }}3x{\text{ }} + 2\cos \left( {3x} \right)\cos \left( x \right)}}{\text{ }}$
On taking the terms common from both the terms
$t = \dfrac{{\sin 3x\left( {{\text{1}} + {\text{ }}2\cos \left( x \right)} \right)}}{{\cos 3x\left( {1 + 2\cos \left( x \right)} \right)}}{\text{ }}$
On cancelling both the terms from numerator and denominator
$t = \dfrac{{\sin 3x}}{{\cos 3x}}{\text{ }}$
We know that $\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$then using this formula
$t = \tan 3x$
Final answer:
The simplest form of the expression $\dfrac{{(\sin {\text{ }}4x{\text{ }} + \sin {\text{ }}3x{\text{ }} + {\text{ }}\sin 2x)}}{{(\cos {\text{ }}4x{\text{ }} + {\text{ }}\cos {\text{ }}3x{\text{ }} + \cos {\text{ }}2x)}}{\text{ }}$ is
$\dfrac{{(\sin {\text{ }}4x{\text{ }} + \sin {\text{ }}3x{\text{ }} + {\text{ }}\sin 2x)}}{{(\cos {\text{ }}4x{\text{ }} + {\text{ }}\cos {\text{ }}3x{\text{ }} + \cos {\text{ }}2x)}}{\text{ }} = {\text{ }}\tan 3x$
So, according to the obtained answer option c is the correct answer.

Note: This question is very tricky we must get idea of using the formula $\sin c + \sin d = 2\sin (\dfrac{{c + d}}{2})\cos (\dfrac{{c - d}}{2})$ by looking the terms ${\text{sin 4x + sin 3x + sin2x}}$. Many students don’t know or may be confused about the formulas. Students commit mistakes in using the formula. They are unable to use the formula properly. They may use the formula on the first two terms or last two terms but from there we are unable to get any common term and not found any canceling the terms from numerator and denominator. And not able to simplify the question properly.