Question

Prove that 4(sin^4x + cos^4x)=cos4x+3

And therefore solve sin^4 + cos^4x=0.5 for 0

Answer 1

Proof:

$4(\sin^4 x + \cos^4 x) =4(\sin^2 x(1-\cos^2 x)+\cos^2x(1-\sin^2x))=\\=4(\sin^2x+\cos^2x-2\sin^2x\cos^2x)=4(1-2\sin^2x\cos^2x)=\\=4(1-\frac{1}{2}\sin^22x)=4(1-\frac{1}{4}(1-\cos4x)=3+\cos4x$

The second part of your question seem truncated in the "for 0" part.

but for the expression to have that value (0.5) the angle should be pi/4 radians.

Answer 2

Trigonometry identities can be proved using several other identities.

• See below for the proof of $\mathbf{4(\sin^4x + \cos^4x) = \cos 4x + 3}$.
• $\mathbf{x=45^o \ or\ \frac{\pi}{4} \ rad}$

Part (a)

The identity is given as:

$\mathbf{4(\sin^4x + \cos^4x) = \cos 4x + 3}$

Expand

$\mathbf{4(\sin^2x \times \sin^2x + \cos^2x \times \cos^2x) = \cos 4x + 3}$

Substitute:

$\mathbf{\sin^2x = 1 - \cos^2x\ and\ \cos^2x = 1 - \sin^2x}$

So, we have:

$\mathbf{4(\sin^2x \times (1 - \cos^2x) + \cos^2x \times (1 - \sin^2x)) = \cos 4x + 3}$

Expand

$\mathbf{4(\sin^2x - \sin^2x\cos^2x + \cos^2x - \cos^2x\sin^2x) = \cos 4x + 3}$

Evaluate like terms

$\mathbf{4(\sin^2x + \cos^2x - 2\cos^2x\sin^2x) = \cos 4x + 3}$

Substitute

$\mathbf{\sin^2x + \cos^2x =1}$

So, we have:

$\mathbf{4(1 - 2\cos^2x\sin^2x) = \cos 4x + 3}$

Substitute

$\mathbf{2\cos^2x\sin^2x =\frac 12\sin^22x}$

So, we have:

$\mathbf{4(1 - \frac 12\sin^22x) = \cos 4x + 3}$

Substitute

$\mathbf{\sin^22x = \frac12(1 - \cos4x)}$

So, we have:

$\mathbf{4(1 - \frac 12( \frac12(1 - \cos4x))) = \cos 4x + 3}$

Expand

$\mathbf{4(1 - \frac 14(1 - \cos4x)) = \cos 4x + 3}$

Expand

$\mathbf{4 -1 + \cos4x = \cos 4x + 3}$

$\mathbf{3 + \cos4x = \cos 4x + 3}$

The identity has been proved

Part (b)

$\mathbf{4(\sin^4x + \cos^4x) = \cos 4x + 3}$

Divide both sides by 4

$\mathbf{\sin^4x + \cos^4x = \frac{\cos 4x + 3}{4}}$

Substitute $\mathbf{\sin^4x + \cos^4x = \frac{\cos 4x + 3}{4}}$ in $\mathbf{ sin^4 + cos^4x=0.5}$

So, we have:

$\mathbf{ \frac{\cos 4x + 3}{4}=0.5}$

Multiply through by 4

$\mathbf{\cos 4x + 3=2}$

Subtract 3 from both sides

$\mathbf{\cos 4x=-1}$

Take arccos of both sides

$\mathbf{x=\cos^{-1}(-1)}$

$\mathbf{4x=180}$

Divide both sides by 4

$\mathbf{x=45}$

Hence, the value of x is:

$\mathbf{x=45^o \ or\ \frac{\pi}{4} \ rad}$

Read more about trigonometry identities at:

https://brainly.com/question/10680548