Chapter 7, Problem 22CT

To determine

To calculate: The exact value of $\sin {75}^{\text{o}}\cos {15}^{\text{o}}$ using sum, difference, product, or half-angle formulas.

Answer to Problem 22CT

Solution:

The exact value of $\sin {75}^{\text{o}}\cos {15}^{\text{o}}$ is $\frac{2+\sqrt{3}}{4}$.

Explanation of Solution

Given information:

The expression is $\sin {75}^{\text{o}}\cos {15}^{\text{o}}$

Formula Used:

Product-to-sum formula: $\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]$

Calculation:

Consider the expression $\sin {75}^{\text{o}}\cos {15}^{\text{o}}$.

Use the Product-to-sum formula $\sin \alpha \cos \beta$ with $\alpha ={75}^{\text{o}}$ and $\beta ={15}^{\text{o}}$

$\Rightarrow \sin {75}^{\text{o}}\cos {15}^{\text{o}}=\frac{1}{2}\left[\sin \left({75}^{\text{o}}+{15}^{\text{o}}\right)+\sin \left({75}^{\text{o}}-{15}^{\text{o}}\right)\right]$

$\Rightarrow \sin {75}^{\text{o}}\cos {15}^{\text{o}}=\frac{1}{2}\left[\sin {90}^{\text{o}}+\sin {60}^{\text{o}}\right]$

By using $\sin {90}^{\text{o}}=1$ and $\sin {60}^{\text{o}}=\frac{\sqrt{3}}{2}$

$\Rightarrow \sin {75}^{\text{o}}\cos {15}^{\text{o}}=\frac{1}{2}\left[1+\frac{\sqrt{3}}{2}\right]$

$\Rightarrow \sin {75}^{\text{o}}\cos {15}^{\text{o}}=\frac{1}{2}\left[\frac{2+\sqrt{3}}{2}\right]$

$\Rightarrow \sin {75}^{\text{o}}\cos {15}^{\text{o}}=\frac{2+\sqrt{3}}{4}$

Therefore, the exact value of $\sin {75}^{\text{o}}\cos {15}^{\text{o}}$ is $\frac{2+\sqrt{3}}{4}$.