Question

1. Use the formula for \(cos\left( {A + B} \right)\) and \(cos\left( {A - B} \right)\) to show that \(sinA\,sinB = \frac{1}{2}\left( {cos\left( {A - B} \right) - cos\left( {A + B} \right)} \right)\).

Short answer

\(sinA\,sinB = \frac{1}{2}\left( {cos\left( {A - B} \right) - cos\left( {A + B} \right)} \right)\)

Step-by-step solution

We know, \(\begin{aligned}{l}cos\left( {A + B} \right) &= cosAcosB - sinAsinB\\cos\left( {A - B} \right) &= cosAcosB + sinAsinB\end{aligned}\)We have to show, \(sinA\,sinB = \frac{1}{2}\left( {cos\left( {A - B} \right) - cos\left( {A + B} \right)} \right)\) Step 1: When we substitute the formula in RHS of the equation and simplify, we get the desired result.

\(\begin{aligned}{l}cos\left( {A + B} \right) &= cosAcosB - sinAsinB\\cos\left( {A - B} \right) &= cosAcosB + sinAsinB\end{aligned}\)

RHS \( = \frac{1}{2}\left( {cos\left( {A - B} \right) - cos\left( {A + B} \right)} \right)\)

RHS \( = \frac{1}{2}\left( {cos A cos B + sin A sin B - \left( {cosAcosB - sinAsinB} \right)} \right)\)

Opening the brackets and simplifying.

RHS \( = \frac{1}{2}\left( {cos A cos B + sin A sin B - cosAcosB + sinAsinB} \right)\)

\( = \frac{1}{2}\left( {2sinAsinB} \right)\)

\( = sinAsinB ....\left( 1 \right)\)

Finalizing the Equation

\(LHS = sinAsinB...\left( 2 \right)\)

From equation 1 and 2

LHS=RHS

Hence, we have proved that \(sinA\,sinB = \frac{1}{2}\left( {cos\left( {A - B} \right) - cos\left( {A + B} \right)} \right)\)

(b). Use part (a) to evaluate \(\int {sin5x\,sin2x\,dx} \).\(\)

Answer:

\(\int {sin5x\,sin2xdx = \frac{1}{6}sin3x - \frac{1}{{14}}sin7x + c} \)

Calculation

\(I = \int {\sin 5x\,\sin 2x\,dx} \)\(\)

Using \(sinA\,sinB = \frac{1}{2}\left( {cos\left( {A - B} \right) - cos\left( {A + B} \right)} \right)\)

\(\begin{aligned}{l}I &= \frac{1}{2}\int {\left( {cos\left( {5x - 2x} \right) - cos\left( {5x + 2x} \right)} \right)} dx\\I &= \frac{1}{2}\int {\left( {cos\left( {3x} \right) - cos\left( {7x} \right)} \right)} dx\\I &= \frac{1}{2}\left( {\int {cos3x\,dx - \int {cos7x\,dx} } } \right)\end{aligned}\)

We know the integration of \(cos\,x = sin\,x\).

Therefore, \(\int {cos\,3x = \frac{{sin3x}}{3}} \) and \(\int {cos\,7x = \frac{{sin7x}}{7}} \)

\(\begin{aligned}{l}I &= \frac{1}{2}\left( {\frac{{sin3x}}{3} - \frac{{sin7x}}{7}} \right) + c\\I &= \frac{{sin3x}}{6} - \frac{{sin7x}}{{14}} + c\end{aligned}\)

Hence, \(I = \int {\sin 5x\sin 2x\,dx = } \frac{1}{6}sin3x - \frac{1}{{14}}sin7x + c\)