Knowing $\sin(45^\circ)={\sqrt2\over2}$, how do I find $\sin(135^\circ)$ and $\cos(135^\circ)$?

Question

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.

Answer 1

Hint:

$135=45+90$. You have formulæ for $\sin(x+90)$ and $\cos(x+90)$.

Answer 2

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, $$\cos(135^\circ) = \cos(180^\circ - 45^\circ) = -\cos(45^\circ)$$