Question

Find the value of $\sin {135}^{\circ }$.

Answer 1

Find the value of the given trigonometric ratio by using the required identity :

Given trigonometric ratio: $\sin {135}^{\circ }$

$\sin {135}^{\circ }$can be expressed as, $\sin {135}^{\circ }=\sin ({90}^{\circ }+{45}^{\circ })$

Using the identity, $$\sin \left(A+B\right)=\sin A\cos B+\cos A\sin B$$ we can write,

$\sin ({90}^{\circ }+{45}^{\circ })=\sin {90}^{\circ }\times \cos {45}^{\circ }+\cos {90}^{\circ }\times \sin {45}^{\circ }$

We know that,
$$\begin{array}{rl}& \sin {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}\\ & \cos {45}^{\circ }={\displaystyle \frac{1}{\sqrt{2}}}\\ & \sin {90}^{\circ }=1\\ & \cos {90}^{\circ }=0\end{array}$$

By substituting the above values we get,

$$\begin{gathered} \sin ({90}^{\circ }+{45}^{\circ })=1\times \frac{1}{\sqrt{2}}+0\times \frac{1}{\sqrt{2}} \\ \Rightarrow \sin \left({135}^{\circ }\right)=\frac{1}{\sqrt{2}} \end{gathered}$$
Hence, the value of $$\sin {135}^{\circ }$$is$${\displaystyle \frac{1}{\sqrt{2}}}$$.