Question

How to find the value of $\sin 4x,\cos 4x,\cot 4x$ ?

Answer 1

First we have to define what the terms we need to solve the problem are. We need to find the value of $\sin 4x,\cos 4x,\cot 4x$. First of all we need to analyse the theta value by separating the theta value as sum identity or multiple identity we can simplify and find the value. We also use basic trigonometric formula to solve basic terms and we also use basic algebraic formula.
Formula to be used:
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
$\cos (A + B) = \cos A\cos B + \sin A\sin B$
$\cot A = \dfrac{1}{{\tan A}}$
$\tan (A + B) = \dfrac{{\tan A + \tan B}}{{1 - \tan A\tan B}}$
$\sin 2A = 2\sin A\cos A$
$\cos 2A = 2{\cos ^2}A - 1$
$\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$
${(a + b)^2} = {a^2} + {b^2} - 2ab$

Complete step by step answer:
To find the value of $\sin 4x$, we use the sum identity of $\sin$,
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
Splitting $4x$ as $2x + 2x$
$\sin 4x$=$\sin (2x + 2x)$
Now applying the sum identity of $\sin$,
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
We know that $A = 2x,B = 2x$
$\sin (2x + 2x) = \sin 2x\cos 2x + \cos 2x\sin 2x$
Adding R.H.S we get,
$\sin (2x + 2x) = 2\sin 2x\cos 2x$
Applying the $\sin 2x = 2\sin x\cos x$and $\cos 2x = 2{\cos ^2}x - 1$ formulae
$\sin 4x = 2(2\sin x\cos x)(2{\cos ^2}x - 1)$
Multiplying we get,
$\sin 4x = 4\sin x\cos x(2{\cos ^2}x - 1)$
Simplifying we get,
$\sin 4x = 8\sin x{\cos ^3}x - 4\sin x\cos x$
Hence, $\sin 4x = 8\sin x{\cos ^3}x - 4\sin x\cos x$
Now, solving $\cos 4x$,
Changing $\cos 4x$as $\cos 2(2x)$
We know that, $\cos 2A = 2{\cos ^2}A - 1$
Considering $A = 2x$,
Substituting the formula in $\cos 4x$,
$\cos 4x = 2{\cos ^2}2x - 1$
Again substituting $\cos 2x$ formula,
We get,
$\cos 4x = 2{(2{\cos ^2}x - 1)^2} - 1$
Substituting ${(a + b)^2} = {a^2} + {b^2} - 2ab$ in the above equation
$\cos 4x = 2({(2{\cos ^2}x)^2} + 1 - 2(2{\cos ^2}x)(1)) - 1$
Simplifying we get,
$\cos 4x = 2(4{\cos ^4}x + 1 - 4{\cos ^2}x) - 1$
$\cos 4x = 8{\cos ^4}x + 2 - 8{\cos ^2}x - 1$
Finally,
$\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1$
Hence, $\cos 4x = 8{\cos ^4}x - 8{\cos ^2}x + 1$
Now, solving $\cot 4x$
We know that, $\cot A = \dfrac{1}{{\tan A}}$
So, writing $\cot 4x$ as $\dfrac{1}{{\tan 4x}}$
Splitting $\tan 4x$as$\tan 2(2x)$,
Then substituting $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$ formula here
We know that $A = 2x$
$\tan 2(2x) = \dfrac{{2\tan 2x}}{{1 - {{\tan }^2}2x}}$
Again using $\tan 2A = \dfrac{{2\tan A}}{{1 - {{\tan }^2}A}}$ formula,
We get,
$\tan 4x = 2\dfrac{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}}}{{1 - {{\left( {\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}} \right)}^2}}}$
Giving the square inside the bracket in denominator
$\tan 4x = 2\dfrac{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}}}{{1 - \dfrac{{2\tan {x^2}}}{{{{(1 - {{\tan }^2}x)}^2}}}}}$
Simplifying denominator,
$\tan 4x = 2\dfrac{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}}}{{\dfrac{{{{(1 - {{\tan }^2}x)}^2} - 2\tan {x^2}}}{{{{(1 - {{\tan }^2}x)}^2}}}}}$
Expanding the denominators’ numerator in ${(a + b)^2} = {a^2} + {b^2} - 2ab$ formula,
$\tan 4x = 2\dfrac{{\dfrac{{2\tan x}}{{1 - {{\tan }^2}x}}}}{{\dfrac{{1 + {{\tan }^4}x - 2{{\tan }^2}x - 2\tan {x^2}}}{{{{(1 - {{\tan }^2}x)}^2}}}}}$
Writing the denominator in multiplication by reciprocal,
$\tan 4x = \dfrac{{4\tan x}}{{1 - {{\tan }^2}x}} \times \dfrac{{{{(1 - {{\tan }^2}x)}^2}}}{{1 + {{\tan }^4}x - 4{{\tan }^2}x}}$
Cancelling $1 - {\tan ^2}x$
$\tan 4x = \dfrac{{4\tan x(1 - {{\tan }^2}x)}}{{1 + {{\tan }^4}x - 4{{\tan }^2}x}}$
Simplifying we get,
$\tan 4x = \dfrac{{4\tan x - 4{{\tan }^2}x}}{{1 + {{\tan }^4}x - 4{{\tan }^2}x}}$
Finally substituting $\tan 4x$in $\cot x$ formula
We know that reciprocal of $\tan 4x$is$\cot x$
$\cot 4x = \dfrac{{1 + {{\tan }^4}x - 4{{\tan }^2}x}}{{4\tan x - 4{{\tan }^2}x}}$
Finally, $\cot 4x = \dfrac{{1 + {{\tan }^4}x - 4{{\tan }^2}x}}{{4\tan x - 4{{\tan }^2}x}}$
Hence found the values.

Note:
We have found the values of $\sin 4x,\cos 4x,\cot 4x$ in one way. But it has more than one way to find the values. For example, we can find $\sin 4x$ by $\sin (3x + x)$ and $\sin 2(2x)$ same thing we can use in $\cos 4x$ also. In $\cot 4x$ we can find the values in terms of $\cot x$ or $\tan x$. In trigonometry we have many ways to find an answer. So, we need not to depend on one way we have many ways to solve it.