Question

Prove that \[\dfrac{{\sin 5x - 2\sin 3x + \sin x}}{{\cos 5x - \cos x}} = \tan x\]

Answer 1

Hint:
In the given function, trigonometric formulas are to be used. Cubic solving it so, for that let us discuss.
Trigonometric functions: - Trigonometric functions are the real functions cubic ratio an angle of right angled triangle to rations of two side lengths.

Complete step by step solution:
Basic trigonometric function:
There are six basic trigonometric functions sine, cosine, tangent, cotangent, Secant and Cosecant. These are related to each other.
For solving the giving question, taking L.H.S. of it.
$\dfrac{{\sin 5x + \sin x - 2\sin 3x}}{{\cos 5x - \cos x}}$
For solving the L.H.S. of the given function, applied trigonometric formulas on it.
The given L.H.S. is
$\dfrac{{(\sin 5x + \sin x) - 2\sin 3x}}{{\cos 5x - \cos x}}$ -------(A)
Firstly solve the numerator of the above equation (A)
$(\sin 5x + \sin x)$ is considered as Numerator
Now, $\sin x + \sin y = 2\sin \dfrac{{x + y}}{2}.\cos \dfrac{{x - y}}{2}$ is the identical term. Apply it on the numerator it will become….
Putting $\sin 5x + \sin x = 2\sin \left( {\dfrac{{5x + x}}{2}} \right).\cos \left( {\dfrac{{5x - x}}{2}} \right)$
$\sin 5x + \sin x = 2\sin \left( {\dfrac{{6x}}{2}} \right).\cos \left( {\dfrac{{4x}}{2}} \right)$
Solving the above term by dividing $2$ we get
$\sin 5x + \sin x = 2\sin 3x\cos 2x$ --------(B)
Solve the Denominator of the equation (A)
$\cos 5x + \cos x$ is the denominator
So,
The identical term is
$\cos x - \cos y = 2\sin \dfrac{{x + y}}{2}.\sin \dfrac{{x - y}}{2}$
According to the given function
$\cos 5x - \cos x$, put $x = 5x$and $y = x$ in identical term we get
$\cos 5x - \cos x = 2\sin \left( {\dfrac{{5x + x}}{2}} \right).\sin \left( {\dfrac{{5x - x}}{2}} \right)$
By adding $5x + x$, get $6x$
By Subtracting $5x - x$, get $4x$ so,
the term becomes…
$\cos 5x - \cos x = 2\sin \left( {\dfrac{{6x}}{2}} \right).\sin \left( {\dfrac{{4x}}{2}} \right)$
Solving the above term
$\dfrac{{6x}}{2}$ will solve to get $3x$
$\dfrac{{4x}}{2}$ will solve to get $2x$
$\cos 5x - \cos x = - 2\sin .3x.\sin 2x$ ………………….(C)
Substituting the value of (B) and (C) i.e. the values of numerator and denominator equation (A)
L.H.S. becomes
$2\sin 3x.\cos 2x = 2\sin 3x - 2\sin 3x\sin 2x$
taking $2\sin 3x$ common from the numerator of about term
$ = \dfrac{{2\sin 3x\left( {\cos 2x - 1} \right)}}{{ - 2\sin 3x \sin 2x}}$
The term $2\sin 3x$ should be cancelled from
$ = \dfrac{{\left( {\cos 2x - 1} \right)}}{{ - \sin 2x}}$
Now, taking $ - 1$ common from the numerator.
$ = \dfrac{{\left( {\cos 2x - 1} \right)}}{{\sin 2x}}$
The $( - )$ minus sign should be cancelled. So the term becomes
$ = \dfrac{{1 - \cos 2x}}{{\sin 2x}}$ -----------(D)
As identical term of $1 - \cos 2x$ is $2{\sin ^2}x$ it will comes from $\cos 2x = 1 - 2{\sin ^2}x$adjusting the turns while taking
$ - 2{\sin ^2}x$ to L.H.S.
 \[\cos 2x + {\sin ^2}x = 1\]
$2{\sin ^2}x = 2\cos x.\sin x$
The identical term $\sin 2x$ will be \[2\cos x.\sin x\]
Substituting the values of both identical terms in equation (D), We get
$ = \dfrac{{1 - \left( {1 - 2{{\sin }^2}x} \right)}}{{2\cos x.\sin x}}$
Solving the above term removes the brackets adjusting the sign.
\[ = \dfrac{{1 - 1 + 2{{\sin }^2}x}}{{2\cos x.\sin x}}\]
As positive $1$ and negative $( - 1)$ becomes zero so,
\[ = \dfrac{{0 + 2{{\sin }^2}x}}{{2\cos x.\sin x}}\]
\[ = \dfrac{{2{{\sin }^2}x}}{{2\cos x.\sin x}}\]
Now, cancelled the term $2\sin x$ from both the numerator and denominator we get -
\[ = \dfrac{{\sin x}}{{\cos x}}\]
as \[\dfrac{{\sin x}}{{\cos x}}\] be the identical term of tan x so,
\[ = \dfrac{{\sin x}}{{\cos x}}\]
\[ = \tan x\]
\[ = R.H.S.\]
Hence the result it proved L.H.S. = R.H.S.

Note:
The above question is solved by using trigonometric functions and formulae. The trigonometric functions are widely used in all science that are related to Geometry. Such as navigations, solid mechanics.