Knowing $\sin(45^\circ)={\sqrt2\over2}$, how do I find $\sin(135^\circ)$ and $\cos(135^\circ)$? 
I just know about $$\sin(45^\circ)={\sqrt2\over2}$$
  and I want understand how to find $\sin(135^\circ)$ and $\cos(135^\circ)$

Thank you.
 A: Hint:
$135=45+90$. You have formulæ for $\sin(x+90)$ and $\cos(x+90)$.
A: The cosine and sine of a directed angle with vertex at the origin and initial side on the positive $x$-axis are, respectively, the $x$- and $y$-coordinates of the points where the terminal side of the angle intersects the unit circle $x^2 + y^2 = 1$.

We are given the sine of a first-quadrant angle.  Since $(\cos\theta, \sin\theta)$ is a point on the unit circle,
$$\cos^2\theta + \sin^2\theta = 1$$
Therefore, we can solve for $\cos(45^\circ)$.
\begin{align*}
\cos^2(45^\circ) + \sin^2(45^\circ) & = 1\\
\cos^2(45^\circ) & = 1 - \sin^2(45^\circ)\\
\cos^2(45^\circ) & = 1 - \left(\frac{\sqrt{2}}{2}\right)^2\\
\cos^2(45^\circ) & = 1 - \frac{2}{4}\\
\cos^2(45^\circ) & = 1 - \frac{1}{2}\\
\cos^2(45^\circ) & = \frac{1}{2}\\
|\cos(45^\circ)| & = \sqrt{\frac{1}{2}}\\
\end{align*}
Since $45^\circ$ is a first-quadrant angle, the $x$-coordinate of the point where the terminal side of the angle intersects the unit circle is positive.  Therefore, its cosine is positive.  Hence, 
\begin{align*}
\cos(45^\circ) & = \sqrt{\frac{1}{2}}\\
               & = \frac{1}{\sqrt{2}}\\
               & = \frac{\sqrt{2}}{2}
\end{align*}
Since $90^\circ < 135^\circ < 180^\circ$, $135^\circ$ is a second-quadrant angle.  To determine its sine and cosine, we can use symmetry.

Since the sine of an angle is the $y$-coordinate of the point where its terminal side intersects the unit circle, two angles that intersect the unit circle at points with the same $y$-coordinate have the same sine.  Hence,
$$\sin(\pi - \theta) = \sin\theta$$
Moreover, two angles with opposite $y$-coordinates have opposite sines.  Hence, 
\begin{align*}
\sin(\pi + \theta) & = -\sin\theta\\
\sin(-\theta) & = -\sin\theta
\end{align*}
Since any two coterminal angles have the same sine, 
$$\sin(\theta + 2k\pi) = \sin\theta, k \in \mathbb{Z}$$
Therefore, $\sin\theta = \sin\varphi$ if 
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or
$$\varphi = \pi - \theta + 2k\pi, k \in \mathbb{Z}$$
If we write $\sin(\pi - \theta) = \sin(\theta)$ in terms of degrees, we obtain 
$$\sin(180^\circ - \theta) = \sin(\theta)$$
Since $135^\circ = 180^\circ - 45^\circ$, we obtain
$$\sin(135^\circ) = \sin(180^\circ - 45^\circ) = \sin(45^\circ)$$
Since the cosine of an angle is the $x$-coordinate of the point where its terminal side intersect the unit circle, two angles that intersect the unit circle at points with the same $x$-coordinate have the same cosine.  Hence,
$$\cos(-\theta) = \cos\theta$$
Moreover, two angles with opposite $x$-coordinates have opposite cosines.  Hence,
\begin{align*}
\cos(\pi - \theta) & = -\cos\theta\\
\cos(\pi + \theta) & = -\cos\theta
\end{align*}
Since any two coterminal angles have the same cosine,
$$\cos(\theta + 2k\pi) = \cos\theta, k \in \mathbb{Z}$$
Therefore, $\cos\theta = \cos\varphi$ if 
$$\varphi = \theta + 2k\pi, k \in \mathbb{Z}$$
or
$$\varphi = -\theta + 2k\pi, k \in \mathbb{Z}$$
If we write $\cos(\pi - \theta) = -\cos\theta$ in terms of degrees, we obtain
$$\cos(180^\circ - \theta) = -\cos\theta$$
Thus,
$$\cos(135^\circ) = \cos(180^\circ - 45^\circ) = -\cos(45^\circ)$$
