Question

How do you find the exact functional value $\sin \left(105°\right)$ using the cosine sum or difference identity?

Answer 1

Solve for the value of $\sin \left(105°\right)$:

Given: $\sin \left(105°\right)$

We know that, $\sin \left(180°-\alpha \right)=\sin \left(\alpha \right)$

And, $\sin \left(105°\right)=\sin \left(180°-75°\right)$
$\begin{array}{cc}=\sin \left(75°\right)& \\ =\sin \left(90-15\right)& \\ =\cos \left(15°\right)& \left[\because \sin \left(90°-\alpha \right)=\cos \left(\alpha \right)\right]\\ =\cos \left(45-30\right)& \end{array}$

using identity $\cos (A-B)=\cos A\cos B+\sin A\cos B$

$\begin{array}{rcl}& & \begin{array}{cc}\Rightarrow & \cos \left(45°-30°\right)=\cos \left(45°\right)\cos \left(30°\right)+\sin \left(45°\right)\sin \left(30°\right)\\ & =\frac{1}{\sqrt{2}}\times \frac{\sqrt{3}}{2}+\frac{1}{\sqrt{2}}\times \frac{1}{2}\\ \Rightarrow & \cos \left(15°\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}\end{array}\end{array}$

But,

$\begin{array}{rcl}& & \begin{array}{cc}& \sin \left(105°\right)=\cos \left(15°\right)\\ \therefore & \sin \left(105°\right)=\frac{\sqrt{3}+1}{2\sqrt{2}}\end{array}\end{array}$

Therefore, $\sin (105°)=\frac{\sqrt{3}+1}{2\sqrt{2}}$.