Question

Find the exact value of:$\sin 45°+\sin 135°+\sin 225°+\sin 315°$

Short answer

The value of$\sin 45°+\sin 135°+\sin 225°+\sin 315°=0$

Step-by-step solution

Step 1. Given information

The given trigonometric function$\sin 45°+\sin 135°+\sin 225°+\sin 315°$

Step 2. Finding the value of sin45°

Recall that the unit circle gives $(x,y)=(\cos \alpha ,\sin \alpha )$ for all the common angles:

Consider$\sin 45°$

Let$\alpha =45°$

From the above unit circle,

The angle$\alpha =45°$ intersects the unit circle at$\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

The pair of coordinates corresponding to $45°$is$\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

Since the second coordinate gives$\sin 45°$

Therefore the value of$\sin 45°=\frac{\sqrt{2}}{2}$ is$\frac{\sqrt{2}}{2}$

step 3. Finding the value of sin135°

Consider$\sin 135°$

Let$\alpha =135°$

From the above unit circle,

The angle$\alpha =135°$ intersects the unit circle at$\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

The pair of coordinates corresponding to $135°$is$\left(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2}\right)$

Since the second coordinate gives$\sin 135°$

Therefore the value of $\sin 135°$is$\frac{\sqrt{2}}{2}$

Step 4. Finding the value of sin225°

Next,consider$\sin 225°$

Let$\alpha =225°$

From the above unit circle,

The anglerole="math" localid="1647676365927" $\alpha =225°$ intersects the unit circle at$\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$

The pair of coordinates corresponding to $225°$is$\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$

Since the second coordinate gives$\sin 225°$

Therefore the value of $\sin 225°$is$-\frac{\sqrt{2}}{2}$

Step 5. Finding the value of sin315°

Consider$\sin 315°$

Let$\alpha =315°$

From the above unit circle,

The angle $\alpha =315°$intersects the unit circle at$\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$

The pair of coordinates corresponding to $315°$is$\left(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2}\right)$

Since the second coordinate gives$\sin 315°$

Therefore the value of $\sin 315°$is$-\frac{\sqrt{2}}{2}$

Step 6. Finding the value of sin45°+sin135°+sin225°+sin315°

Finally add all these four exact values to obtain the required solution

Therefore,

$$\begin{gathered} \sin 45°+\sin 135°+\sin 225°+\sin 315°=\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}+\left(-\frac{\sqrt{2}}{2}\right)+\left(-\frac{\sqrt{2}}{2}\right) \\ =0 \end{gathered}$$

Hence answer is$0$