Derivative of y = cos(4x)
Derivative of y = cos(4x)

In this blog post, we will learn how to simplify the trigonometric expressions cos4 x -sin4 x and cos4x+sin4x. We will apply the following two trigonometric identities:

• $\cos^2x+\sin^2x=1$ $\cdots (\star)$
• $\cos^2x-\sin^2x=\cos 2x$ $\cdots (\star \star)$
• $2\cos x\sin x=\sin 2x$ $\cdots (\star \star \star)$

## cos4x-sin4x Formula

To simplify the expression cos4 x -sin4 x, we first apply the formula a2-b2 =(a-b)(a+b) with $a=\cos^2x$ and $b=\sin^2x$.

Therefore, $\cos^4x-\sin^4x$ $=(\cos^2x)^2 – (\sin^2 x)^2$

$=(\cos^2x-\sin^2x)(\cos^2x+\sin^2x)$

$=\cos 2x \cdot 1$ by the above formulas $(\star)$ and $(\star \star)$.

$=\cos 2x$

So the formula of cos4 x -sin4 x is given as follows: cos4 x -sin4 x=cos 2x.

sinx=0, cosx=0, tanx=0 General Solution

Values of sin 15, cos 15, tan 15

Values of sin 75, cos 75, tan 75

## cos4x+sin4x Formula

To simplify the expression cos4x+sin4x, we first apply the formula a2+b2 =(a+b)2-2ab with $a=\cos^2x$ and $b=\sin^2x$. Then we have

$\cos^4x+\sin^4x$

$=(\cos^2x)^2 + (\sin^2 x)^2$

$=(\cos^2x+\sin^2x)^2 -2\cos^2x\sin^2x$

$=(1)^2 -2\cos^2x\sin^2x$ by the above formula $(\star)$.

$=1-\dfrac{1}{2} (4\cos^2x \sin^2x)$

$=1-\dfrac{1}{2} (2\cos x \sin x)^2$

$=1-\dfrac{1}{2} (\sin 2x)^2$ by the above formula $(\star \star \star)$

$=1-\dfrac{\sin^2 2x}{2}$

So the formula of cos4x+sin4x is given as follows: cos4x+sin4x = $1-\dfrac{\sin^2 2x}{2}$.

## FAQs

Q1: What is the formula of cos4x+sin4x?

Answer: cos4x+sin4x = $1-\dfrac{\sin^2 2x}{2}$.

Q2: What is the formula of cos4x-sin4x?

Answer: cos4 x -sin4 x=cos 2x.

You are watching: Simplify cos^4x+sin^4x – iMath. Info created by Bút Chì Xanh selection and synthesis along with other related topics.