$\dfrac{\cos{x}-\cos{3x}}{\cos{x}}$ $+$ $\dfrac{\sin{x}+\sin{3x}}{\sin{x}}$
$=\,\,$ $\dfrac{2\sin{\bigg(\dfrac{3x+x}{2}\bigg)}\sin{\bigg(\dfrac{3x-x}{2}\bigg)}}{\cos{x}}$ $+$ $\dfrac{2\sin{\bigg(\dfrac{3x+x}{2}\bigg)}\cos{\bigg(\dfrac{3x-x}{2}\bigg)}}{\sin{x}}$
There is nothing to simplify in the denominator of both terms in the trigonometric expression. So, let’s focus on simplifying the numerator of every term in the trigonometric expression.
$=\,\,$ $\dfrac{2\sin{\bigg(\dfrac{4x}{2}\bigg)}\sin{\bigg(\dfrac{2x}{2}\bigg)}}{\cos{x}}$ $+$ $\dfrac{2\sin{\bigg(\dfrac{4x}{2}\bigg)}\cos{\bigg(\dfrac{2x}{2}\bigg)}}{\sin{x}}$
$=\,\,$ $\dfrac{2\sin{\bigg(\dfrac{\cancel{4}x}{\cancel{2}}\bigg)}\sin{\bigg(\dfrac{\cancel{2}x}{\cancel{2}}\bigg)}}{\cos{x}}$ $+$ $\dfrac{2\sin{\bigg(\dfrac{\cancel{4}x}{\cancel{2}}\bigg)}\cos{\bigg(\dfrac{\cancel{2}x}{\cancel{2}}\bigg)}}{\sin{x}}$
$=\,\,$ $\dfrac{2\sin{(2x)}\sin{(x)}}{\cos{x}}$ $+$ $\dfrac{2\sin{(2x)}\cos{(x)}}{\sin{x}}$
$=\,\,$ $\dfrac{2\sin{2x}\sin{x}}{\cos{x}}$ $+$ $\dfrac{2\sin{2x}\cos{x}}{\sin{x}}$
The numerator in both terms consists of sine function with double angle. So, the sine of two times angle $x$ can be expanded by the double angle identity of sine function.
$=\,\,$ $\dfrac{2 \times 2\sin{x}\cos{x} \times \sin{x}}{\cos{x}}$ $+$ $\dfrac{2 \times 2\sin{x}\cos{x} \times \cos{x}}{\sin{x}}$
$=\,\,$ $\dfrac{2 \times 2\sin{x}\cancel{\cos{x}} \times \sin{x}}{\cancel{\cos{x}}}$ $+$ $\dfrac{2 \times 2\cancel{\sin{x}}\cos{x} \times \cos{x}}{\cancel{\sin{x}}}$
$=\,\,$ $2 \times 2\sin{x} \times \sin{x}$ $+$ $2 \times 2\cos{x} \times \cos{x}$
The expressions in both terms of the trigonometric expression are released from rational form. Now, simplify each term by multiplying the factors in the expression.
$=\,\,$ $4\sin^2{x}$ $+$ $4\cos^2{x}$
The number $4$ is a common factor in both terms of the trigonometric expression. It can be taken out common from them to simplify it further.
$=\,\,$ $4 \times \big(\sin^2{x}+\cos^2{x}\big)$
The sum of squares of sine and cosine functions is equal to one as per the Pythagorean identity of sine and cosine.
$=\,\,$ $4 \times (1)$
$=\,\,$ $4 \times 1$
$=\,\,$ $4$
$\dfrac{\cos{x}-\cos{3x}}{\cos{x}}$ $+$ $\dfrac{\sin{x}+\sin{3x}}{\sin{x}}$
Learn how to find the $\cos{x}$ minus $\cos{3x}$ divided by $\cos{x}$ plus $\sin{x}$ plus $\sin{3x}$ divided by $\sin{x}$ by the triple angle trigonometric identities.
A best free mathematics education website for students, teachers and researchers.
Learn each topic of the mathematics easily with understandable proofs and visual animation graphics.
Learn how to solve the maths problems in different methods with understandable steps.
Copyright © 2012 - 2022 Math Doubts, All Rights Reserved