Q. 85 - Precalculus Enhanced with Graphing Utilities

Question

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

Step-by-Step Solution

Answer

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

Step 1. Given information

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

Step 2. Finding the value of sin 45 °

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 sin 135 °

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 sin 225 °

Next,consider$\sin 225°$

Let $\alpha =225°$

From the above unit circle,

The angle  $\alpha =225°$ 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 sin 315 °

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 sin 45 ° + sin 135 ° + sin 225 ° + sin 315 °

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 $0$