Question

: How do you express $\sin \left( \dfrac{\pi }{12} \right)\cdot \cos \left( \dfrac{3\pi }{8} \right)$ without products of trigonometric functions?

Answer 1

Simplify $\sin \left( \dfrac{\pi }{12} \right)$ and $\cos \left( \dfrac{3\pi }{8} \right)$ using the half angle identity for sine function and cosine function separately. Then multiply the values together and do necessary calculations to obtain the required solution.

Complete step by step solution:
According to the given question, we are given an expression which has the product of two trigonometric functions. We are asked to express it separately.
Here, the question says that we have to express it without products of trigonometric functions, that simply means that, we have to express the functions separately.
We will accomplish this task by making use of a trigonometric formula, which is,
$\sin (a-b)+\sin (a+b)=2\sin a\cos b$
We can rearrange it as,
$\sin a\cos b=\dfrac{1}{2}\left( \sin (a-b)+\sin (a+b) \right)$------(1)
The given expression we have is,
$\sin \left( \dfrac{\pi }{12} \right).\cos \left( \dfrac{3\pi }{8} \right)$-----(2)
Here, $a=\dfrac{\pi }{12}$ and $b=\dfrac{3\pi }{8}$
We will obtain the values and substitute it in the equation (1),
We now have,
$\sin (a-b)=\sin \left( \dfrac{\pi }{12}-\dfrac{3\pi }{8} \right)$
Taking LCM, we have, $LCM(12,8)=24$
$\Rightarrow \sin \left( \dfrac{2\pi -9\pi }{24} \right)=\sin \left( \dfrac{-7\pi }{24} \right)$
And as we know that sine function is an odd function, so we have,
$\Rightarrow -\sin \left( \dfrac{7\pi }{24} \right)$-----(3)
Now, then,
$\sin (a+b)=\sin \left( \dfrac{\pi }{12}+\dfrac{3\pi }{8} \right)$
$\Rightarrow \sin \left( \dfrac{2\pi +9\pi }{24} \right)$
$\Rightarrow \sin \left( \dfrac{11\pi }{24} \right)$----(4)
Now, we will substitute the equation (3) and equation (4) in equation (1), we get,
$\sin \left( \dfrac{\pi }{12} \right)\cos \left( \dfrac{3\pi }{8} \right)=\dfrac{1}{2}\left( -\sin \left( \dfrac{7\pi }{24} \right)+\sin \left( \dfrac{11\pi }{24} \right) \right)$-----(5)
We can move a step ahead and determine the numerical value as well. We will have to figure out the values of the terms involved first.
We know that,
$\sin \theta =\cos ({{90}^{\circ }}-\theta )$
We can write the equation (5) as,
$\dfrac{1}{2}\left( -\sin \left( \dfrac{7\pi }{24} \right)+\sin \left( \dfrac{11\pi }{24} \right) \right)$
$\Rightarrow \dfrac{1}{2}\left( -\cos \left( \dfrac{\pi }{2}-\dfrac{7\pi }{24} \right)+\cos \left( \dfrac{\pi }{2}-\dfrac{11\pi }{24} \right) \right)$
$\Rightarrow \dfrac{1}{2}\left( -\cos \left( \dfrac{5\pi }{24} \right)+\cos \left( \dfrac{\pi }{24} \right) \right)$-----(6)
So, we have to find the value of $\cos \left( \dfrac{5\pi }{24} \right)$ and $\cos \left( \dfrac{\pi }{24} \right)$. We will be using formula $\cos \left( \dfrac{A}{2} \right)=\sqrt{\dfrac{1+\cos A}{2}}$ for the same.
To find the value of $\cos \left( \dfrac{5\pi }{24} \right)$, we will first find the value of $\cos \left( \dfrac{5\pi }{12} \right)$.
$\cos \left( \dfrac{5\pi }{12} \right)=\cos \left( \dfrac{3\pi +2\pi }{12} \right)=\cos \left( \dfrac{3\pi }{12}+\dfrac{2\pi }{12} \right)=\cos \left( \dfrac{\pi }{4}+\dfrac{\pi }{6} \right)$
We will use the formula $\cos (A+B)=\cos A\cos B-\sin A\sin B$, we get,
$\Rightarrow \cos \dfrac{\pi }{4}\cos \dfrac{\pi }{6}-\sin \dfrac{\pi }{4}\sin \dfrac{\pi }{6}$
$\Rightarrow \dfrac{1}{\sqrt{2}}.\dfrac{\sqrt{3}}{2}-\dfrac{1}{\sqrt{2}}.\dfrac{1}{2}=\dfrac{\sqrt{3}-1}{2\sqrt{2}}.\dfrac{\sqrt{2}}{\sqrt{2}}$
$\Rightarrow \dfrac{\sqrt{6}-\sqrt{2}}{4}$
So,
$\cos \left( \dfrac{5\pi }{24} \right)=\sqrt{\dfrac{1+\dfrac{\sqrt{6}-\sqrt{2}}{4}}{2}}$
$\cos \left( \dfrac{5\pi }{24} \right)=\sqrt{\dfrac{4+\sqrt{6}-\sqrt{2}}{8}}=\dfrac{\sqrt{4+\sqrt{6}-\sqrt{2}}}{2\sqrt{2}}$-------(7)
Now, we will find the value of $\cos \left( \dfrac{\pi }{24} \right)$, but first we will find the value of $\cos \left( \dfrac{\pi }{12} \right)$,
Similarly, $\cos \left( \dfrac{\pi }{12} \right)=\cos \left( \dfrac{\pi }{3}-\dfrac{\pi }{4} \right)$
Using the formula $\cos (A+B)=\cos A\cos B+\sin A\sin B$, we have,
$\Rightarrow \cos \dfrac{\pi }{4}\cos \dfrac{\pi }{3}+\sin \dfrac{\pi }{4}\sin \dfrac{\pi }{3}$
$\Rightarrow \dfrac{1}{\sqrt{2}}.\dfrac{1}{2}+\dfrac{1}{\sqrt{2}}.\dfrac{\sqrt{3}}{2}=\dfrac{1+\sqrt{3}}{2\sqrt{2}}.\dfrac{\sqrt{2}}{\sqrt{2}}$
$\Rightarrow \dfrac{\sqrt{6}+\sqrt{2}}{4}$
$\cos \left( \dfrac{\pi }{24} \right)=\sqrt{\dfrac{1+\dfrac{\sqrt{6}+\sqrt{2}}{4}}{2}}$
$\cos \left( \dfrac{\pi }{24} \right)=\sqrt{\dfrac{4+\sqrt{6}+\sqrt{2}}{8}}=\dfrac{\sqrt{4+\sqrt{6}+\sqrt{2}}}{2\sqrt{2}}$------(8)
Now, we will put equation (7) and (8) in equation (6), we get,
$\Rightarrow \dfrac{1}{2}\left( -\dfrac{\sqrt{4+\sqrt{6}-\sqrt{2}}}{2\sqrt{2}}+\dfrac{\sqrt{4+\sqrt{6}+\sqrt{2}}}{2\sqrt{2}} \right)$
$\Rightarrow \dfrac{1}{2}\left( \dfrac{\sqrt{4+\sqrt{6}+\sqrt{2}}-\sqrt{4+\sqrt{6}-\sqrt{2}}}{2\sqrt{2}} \right)$
$\Rightarrow \left( \dfrac{\sqrt{4+\sqrt{6}+\sqrt{2}}-\sqrt{4+\sqrt{6}-\sqrt{2}}}{4\sqrt{2}} \right)$
$\sin \left( \dfrac{\pi }{12} \right)\cos \left( \dfrac{3\pi }{8} \right)=\left( \dfrac{\sqrt{4+\sqrt{6}+\sqrt{2}}-\sqrt{4+\sqrt{6}-\sqrt{2}}}{4\sqrt{2}} \right)$------(9)
Therefore, the expression we have is,
$\sin \left( \dfrac{\pi }{12} \right)\cos \left( \dfrac{3\pi }{8} \right)=\dfrac{1}{2}\left( -\sin \left( \dfrac{7\pi }{24} \right)+\sin \left( \dfrac{11\pi }{24} \right) \right)$.
And the numerical value of the given expression is,
$\sin \left( \dfrac{\pi }{12} \right)\cos \left( \dfrac{3\pi }{8} \right)=\left( \dfrac{\sqrt{4+\sqrt{6}+\sqrt{2}}-\sqrt{4+\sqrt{6}-\sqrt{2}}}{4\sqrt{2}} \right)$.

Note: The angles $\sin \left( \dfrac{\pi }{12} \right)$ and $\cos \left( \dfrac{3\pi }{8} \right)$ should be simplified separately to avoid complex calculations and errors. Sign convention for different trigonometric functions should be taken properly as per the quadrant. Some basic values of sine and cosine function should be remembered for faster and exact calculations.