# Question d76e9

Jul 24, 2017

$0$

#### Explanation:

$\sin 2 x = 1$, $2 x = \frac{\pi}{2}$ or $x = \frac{\pi}{4}$

Hence

$u = \cos \left(8 \cdot \frac{\pi}{4}\right) + 2 \cos \left(6 \cdot \frac{\pi}{4}\right) + \cos \left(4 \cdot \frac{\pi}{4}\right)$

$= \cos \left(2 \pi\right) + 2 \cos \left(\frac{3 \pi}{2}\right) + \cos \left(\pi\right)$

$= 1 + 2 \cdot 0 - 1$

$= 0$

1) I found $x$ for $\sin 2 x = 1$ condition.

2) I substituted to $x = \frac{\pi}{4}$.

Jul 24, 2017

$0$.

#### Explanation:

$\sin \left(2 x\right) = 1$.

In general, equations of the form $\sin \left(\theta\right) = K$ and solved by $\theta = \arcsin \left(K\right) + 2 n \pi$ and $\pi - \theta = \arcsin \left(K\right) + 2 n \pi$, $n \in \mathbb{Z}$, because of the periodicity of the sine function.

Note that $\arcsin \left(1\right) = \frac{\pi}{2}$. This means that if $\sin \left(2 x\right) = 1$,

$2 x = \frac{\pi}{2} + 2 n \pi$,
$x = \frac{\pi}{4} + n \pi$.

$\pi - 2 x = \frac{\pi}{2} + 2 n \pi$,
$2 x = \frac{\pi}{2} - 2 n \pi$,
$x = \frac{\pi}{4} - n \pi$.

As $n \in \mathbb{Z}$ these solution sets are identical.
So if $\sin \left(2 x\right) = 1$ then $x = \frac{\pi}{4} + n \pi$, $n \in \mathbb{Z}$.

Then

cos(8(pi/4+npi)+2cos(6(pi/4+npi))+cos(4(pi/4+npi)),
$\cos \left(2 \pi + 4 \left(2 n \pi\right)\right) + 2 \cos \left(\frac{3 \pi}{2} + 3 \left(2 n \pi\right)\right) + \cos \left(\pi + 2 \left(2 n \pi\right)\right)$.

Multiplies of the periodicity can be removed and the values substituted to give,

$1 + 2 \cdot 0 - 1 = 0$.

In this case, it was safe to assume $\sin \left(2 x\right) = 1 \implies x = \frac{\pi}{4}$. This will not always be the case! So it is good to solve more generally.