# trigonometric identity

#### matemagia

Can you prove that:
$$\displaystyle (7+\cos (4x))^2=32[1+\sin^8(x)+\cos^8(x)]$$?
Help needed.

#### MarkFL

Re: trigonometric identity

Starting with the left side, let's first employ a double angle identity for cosine:

$$\displaystyle (7+\cos(4x))^2=(6+2\cos^2(2x))^2=4(3+\cos^2(2x))^2$$

Now, since we need both the sine and cosine functions in the final result, let's use a different double-angle identity for cosine:

$$\displaystyle 4(3+(\cos^2(x)-\sin^2(x))^2)^2=4(3+\cos^4(x)-2\cos^2(x)\sin^2(x)+\sin^4(x))^2$$

Now, let's look at the term:

$$\displaystyle 2\cos^2(x)\sin^2(x)=\cos^2(x)\sin^2(x)+\cos^2(x)\sin^2(x)=\cos^2(x)(1-\cos^2(x)) + (1-\sin^2(x))\sin^2(x)=$$

$$\displaystyle \cos^2-\cos^4(x)+\sin^2(x)-\sin^4(x) = 1-(\cos^4(x)+\sin^4(x))$$

Now, substitute this into the above:

$$\displaystyle 4(3+\cos^4(x)-(1-(\cos^4(x)+\sin^4(x)))+\sin^4(x))^2 = 16(1+\cos^4(x)+\sin^4(x))^2$$

Expanding the squared term, we have:

$$\displaystyle 16(1+\sin^8(x)+\cos^8(x)+2\cos^4(x)+2\sin^4(x)+2\sin^4(x)\cos^4(x))$$

Now, observe, we need to prove:

$$\displaystyle 2\cos^4(x)+2\sin^4(x)+2\sin^4(x)\cos^4(x) = 1+\sin^8(x)+\cos^8(x)$$

$$\displaystyle 2\cos^4(x)+2\sin^4(x)-1 = \sin^8(x)-2\sin^4(x)\cos^4(x)+\cos^8(x)$$

$$\displaystyle 2\cos^4(x)+2\sin^4(x)-1 = (\sin^4(x)-\cos^4(x))^2$$

$$\displaystyle 2\cos^4(x)+2\sin^4(x)-1 = ((\sin^2(x)+\cos^2(x))(\sin^2(x)-\cos^2(x)))^2$$

$$\displaystyle 2\cos^4(x)+2\sin^4(x)-1 = (\sin^2(x)-\cos^2(x))^2$$

$$\displaystyle 2\cos^4(x)+2\sin^4(x)-1 = \sin^4(x)-2\sin^2(x)\cos^2(x)+\cos^4(x)$$

$$\displaystyle \cos^4(x)+\sin^4(x)-1 = -2\sin^2(x)\cos^2(x)$$

Recall, we earlier found this identity, thus, we may now state:

$$\displaystyle (7+\cos(4x))^2 = 16(2+2\sin^8(x)+2\cos^8(x)) = 32(1+\sin^8(x)+\cos^8(x))$$

#### matemagia

Thanks! Respect! MarkFL

#### soroban

Math Team
Hello, mathmagia!

I proved it "the other way".

$$\displaystyle \text{Prove that: }\: (7\,+\,\cos 4x)^2\: = \:32[1\,+\,\sin^8x\,+\,\cos^8x]$$

We will use these identities: .$$\displaystyle \begin{Bmatrix}\sin^2\theta &=& \displaystyle\frac{1-\cos2\theta}{2} \\ \\ \\\cos^2\!\theta &=& \displaystyle\frac{1 + \cos2\theta}{2}\end{Bmatrix}$$

The right side is: .$$\displaystyle 32\left[1\,+\,\sin^8x\,+\,\cos^8x\right]$$

. . . $$\displaystyle =\;32\left[1\,+\,(\sin^2x)^4\,+\,(\cos^2x)^4\right] \;=\;32\left[1\,+\,\left(\frac{1\,-\,\cos2x}{2}\right)^4\,+\,\left(\frac{1\,+\,\cos2x}{2}\right)^4\right]$$

. . . $$\displaystyle =\;32\left[1\,+\,\frac{1\,-\,4\cos2x\,+\,6\cos^22x\,-\,4\cos^32x\,+\,\cos^42x}{16}\,+\,\frac{1\,+\,4\cos^22x\,+\,6\cos^22x\,+\,4\cos^32x\,+\,\cos^42x}{16}\right]$$

. . . $$\displaystyle =\;32\left[1\,+\,\frac{2\,+\,12\cos^22x\,+\,2\cos^42x}{16}\right] \;=\;32\left[\frac{8\,+\,1\,+\,6\cos^22x\,+\,\cos^42x}{8}\right]$$

. . . $$\displaystyle =\;4\left[9\,+\,6\cos^22x\,+\,\cos^42x\right] \;=\;4\left[9\,+\,6\left(\frac{1\,+\,\cos4x}{2}\right)\,+\,(\cos^22x)^2\right]$$

. . . $$\displaystyle =\;4\left[9\,+\,3(1\,+\,\cos4x)\,+\, \left(\frac{1\,+\,\cos4x}{2}\right)^2\right] \;=\;4\left[9\,+3\,+\,3\cos4x + \frac{1\,+\,2\cos4x\,+\,\cos^24x}{4}\right]$$

. . . $$\displaystyle =\;4\left[\frac{48\,+\,12\cos4x\,+\,1\,+\,2\cos4x\,+\,\cos^24x}{4}\right] \;=\;49\,+\,14\cos4x\,+\,\cos^24x$$

. . . $$\displaystyle =\;(7\,+\,\cos4x)^2$$

#### MarkFL

Another method (very similar to my first), beginning from the left side:

$$\displaystyle (7+\cos(4x))^2 = (7+\cos(3x+x))^2 = (7 + \cos(3x)\cos(x) - \sin(3x)\sin(x))^2=$$

$$\displaystyle \left(7+4\cos^4(x)-3\cos^2(x)-3\sin^2(x)+4\sin^4(x)\right)^2=16\left(1+\sin^4(x)+\cos^4(x)\right)^2$$

The rest follows as in my first post.

#### skipjack

Forum Staff
$(7 + \cos(4x))^2 = (2\cos^2(4x) + 28\cos(4x) + 98)/2 = (\cos(8x) + 28\cos(4x) + 99)/2$.

Using the identities
$\cos^8(x) = (\cos(8x) + 8\cos(6x) + 28\cos(4x) + 56\cos(2x) + 35)/128$ and
$\sin^8(x) = (\cos(8x) - 8\cos(6x) + 28\cos(4x) - 56\cos(2x) + 35)/128$,

$32(\cos^8(x) + \sin^8(x) + 1) = (\cos(8x) + 28\cos(4x) + 99)/2$.

greg1313