Question

What is $\dfrac{{\cos 7x - \cos 3x}}{{\sin 7x - 2\sin 5x + \sin 3x}}$equal to?
A) $\tan x$
B) $\cot x$
C) $\tan 2x$
D) $\cot 2x$

Answer 1

Hint: This is a problem based on trigonometric function. Initially we will use the $cosA \pm cosB$ and $sinA\pm sinB$ formula to simplify it. Then we will use some other trigonometric formulas to get our required answer. Some of the formulas are mentioned below.
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}$
$\sin 2\theta = 2\sin \theta \cos \theta $
$\cos 2\theta = 2{\cos ^2}\theta - 1 = 1 - 2{\sin ^2}\theta $
$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\sin A + \sin B = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$

Complete answer:
The given trigonometric function is $\dfrac{{\cos 7x - \cos 3x}}{{\sin 7x - 2\sin 5x + \sin 3x}}$. The objective is to solve the given function using trigonometric identities.
First, simplify the numerator term of the given function by using the identity,
$\cos C - \cos D = - 2\sin \left( {\dfrac{{C + D}}{2}} \right)\sin \left( {\dfrac{{C - D}}{2}} \right)$
\[\dfrac{{ - 2\sin \left( {\dfrac{{7x + 3x}}{2}} \right)\sin \left( {\dfrac{{7x - 3x}}{2}} \right)}}{{\sin 7x - 2\sin 5x + \sin 3x}}\]
\[ = \dfrac{{ - 2\sin \left( {\dfrac{{10x}}{2}} \right)\sin \left( {\dfrac{{4x}}{2}} \right)}}{{\sin 7x - 2\sin 5x + \sin 3x}}\]
\[ = \dfrac{{ - 2\sin 5x\sin 2x}}{{\sin 7x - 2\sin 5x + \sin 3x}}\]
Now, we need to solve the denominator terms, use the identity given by
$\sin C + \sin D = 2\sin \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$
To evaluate the terms $\sin 7x + \sin 3x$
\[ = \dfrac{{ - 2\sin 5x\sin 2x}}{{2\sin \left( {\dfrac{{7x + 3x}}{2}} \right)\cos \left( {\dfrac{{7x - 3x}}{2}} \right) - 2\sin 5x}}\]
\[ = \dfrac{{ - 2\sin 5x\sin 2x}}{{2\sin \left( {\dfrac{{10x}}{2}} \right)\cos \left( {\dfrac{{4x}}{2}} \right) - 2\sin 5x}}\]
\[ = \dfrac{{ - 2\sin 5x\sin 2x}}{{2\sin 5x\cos 2x - 2\sin 5x}}\]
Take $ - 2\sin 5x$ common from both numerator and denominator
\[ = \dfrac{{ - 2\sin 5x(\sin 2x)}}{{2\sin 5x(1 - \cos 2x)}}\]
Cancel the term $ - 2\sin 5x$ from both the numerator and denominator, to get
\[ = \dfrac{{\sin 2x}}{{1 - \cos 2x}}\]
Use the identity given by: $\sin 2x = 2\sin x\cos x$ to evaluate the numerator term of the given trigonometric function,
\[ = \dfrac{{2\sin x\cos x}}{{1 - \cos 2x}}\]
Now use the identity given by:$\cos 2x = 1 - 2{\sin ^2}x$ to replace the term $\cos 2x$in the denominator,
\[ = \dfrac{{2\sin x\cos x}}{{1 - (1 - 2{{\sin }^2}x)}}\]
Open the bracket term in the denominator by reversing the sign inside the bracket
\[ = \dfrac{{2\sin x\cos x}}{{1 - 1 + 2{{\sin }^2}x}}\]
\[ = \dfrac{{2\sin x\cos x}}{{2{{\sin }^2}x}}\]
Take the term $2\sin x$ common from the numerator and denominator
\[ = \dfrac{{2\sin x(\cos x)}}{{2\sin x(\sin x)}}\]
Cancel the term $2\sin x$from the numerator and denominator
\[ = \dfrac{{\cos x}}{{\sin x}}\]
Since, we know that $\cot x = \dfrac{{\cos x}}{{\sin x}}$, so
$ = \cot x$
Hence, we get that $\dfrac{{\cos 7x - \cos 3x}}{{\sin 7x - 2\sin 5x + \sin 3x}} = \cot x$.

Therefore, the correct option is B

Note: Trigonometric functions can be simply defined as the functions of an angle of a triangle and the basic trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. Also, whenever we are asked to calculate or solve the given trigonometric expression or equation, we should be able to apply the appropriate trigonometric identities and formulae. Sometimes we tend to apply the algebraic identities too.