Question

If $\sin \theta =\sin \left(15°\right)+\sin \left(45°\right)$, where $0°<\theta <90°$, then $\theta$ is equal to

• A. $45°$
• B. $54°$
• C. $60°$
• D. $72°$
• E. $75°$

Answer 1

The correct option is E

$75°$

Explanation for the correct option

Step 1: Explanation of the formula required for the solution

Suppose there are two angles $A\text{and}B$, then the formula for $\sin \left(\mathrm{A}\right)+\sin \left(\mathrm{B}\right)$ is given as

$\mathbf{sin}\mathbf{\left(}\mathbf{A}\mathbf{\right)}\mathbf{+}\mathbf{sin}\mathbf{\left(}\mathbf{B}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{sin}\mathbf{\left(}\frac{\mathbf{A}\mathbf{+}\mathbf{B}}{\mathbf{2}}\mathbf{\right)}\mathit{c}\mathit{o}\mathit{s}\mathbf{\left(}\frac{\mathbf{A}\mathbf{-}\mathbf{B}}{\mathbf{2}}\mathbf{\right)}$

For this problem, $A=15°\text{and}B=45°$.

Step 2: calculate the value of $\mathit{\theta }$

Apply the above formula in the given,

$\begin{array}{rcl}\therefore \sin \theta & =& \sin \left(15°\right)+\sin \left(45°\right)\\ & =& 2\sin \left(\frac{15°+45°}{2}\right)\cos \left(\frac{15°-45°}{2}\right)\\ & =& 2\sin \left(\frac{60°}{2}\right)\cos \left(\frac{-30°}{2}\right)\\ & =& 2\sin \left(30°\right)\cos \left(-15°\right)\end{array}$

Due to the given condition $0°<\theta <90°$ we have

$\cos (-15°)=\cos (15°)$

Because every trigonometric function in the first quadrant is positive.

Now, we know that $\sin \left(30°\right)=\frac{1}{2}$,

$\begin{array}{rcl}\therefore \sin \left(\theta \right)& =& 2\left(\frac{1}{2}\right)\cos \left(15°\right)\\ & =& \cos \left(15°\right)\end{array}$

Step 3: Apply the concept $\mathbf{cos}\left(\frac{\pi }{2}-\alpha \right)\mathbf{=}\mathbf{sin}\left(\alpha \right)$, then we have

$\begin{array}{rcl}\therefore \sin \left(\theta \right)& =& \cos \left(90°-75°\right)\\ & =& \sin \left(75°\right)\end{array}$

Hence, the correct option is (E).