$\dfrac{\cos{x}-\cos{3x}}{\cos{x}}$ $+$ $\dfrac{\sin{x}+\sin{3x}}{\sin{x}}$

$=\,\,$ $\dfrac{2\sin{\bigg(\dfrac{3x+x}{2}\bigg)}\sin{\bigg(\dfrac{3x-x}{2}\bigg)}}{\cos{x}}$ $+$ $\dfrac{2\sin{\bigg(\dfrac{3x+x}{2}\bigg)}\cos{\bigg(\dfrac{3x-x}{2}\bigg)}}{\sin{x}}$

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.

Latest Math Topics

Jul 20, 2023

Jun 26, 2023

Jun 23, 2023

Latest Math Problems

Oct 15, 2023

Copyright © 2012 - 2023 Math Doubts, All Rights Reserved