Question

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

Short answer

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

The formula for,

\(\begin{aligned}{l}sin\left( {A + B} \right) = sinAcosB - cosAsinB\\sin\left( {A - B} \right) = sinAcosB + cosAsinB\end{aligned}\)

We have to show,

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

Step-by-step solution

When we substitute the formula in RHS of the equation and simplify, we get the desired result.

\(\begin{aligned}{l}sin\left( {A + B} \right) &= sinAcosB - cosAsinB\\sin\left( {A - B} \right) &= sinAcosB + cosAsinB\end{aligned}\)

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

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

Simplifying the above equation.

\( = \frac{1}{2}\left( {2sinA\,cosB} \right)\)

\( = sinA\,cos\,B ....\left( 1 \right)\)

\(LHS = sinA\,cosB...\left( 2 \right)\)

From equation 1 and 2

LHS=RHS

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

(b). Use part (a) to evaluate \(\int {sin3x\,cosx\,dx} \).\(\)

Answer:

\(\int {sin3x\,cosxdx = - \frac{1}{4}cos2x - \frac{1}{8}cos 4x + c} \)

In the formula, assume\(A = 3x\)and\(B = x\)Substitute these values in the formula and simplify. Finally the integral gets simplified to evaluate.\(\)

Step 1:

\(I = \int {sin3x\,cosx\,dx} \)\(\)

Let \(A = 3x\) and \(B = x\).

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

\(\begin{aligned}{l}I &= \frac{1}{2}\int {\left( {sin\left( {3x - x} \right) - sin\left( {3x + x} \right)} \right)} dx\\I &= \frac{1}{2}\int {\left( {sin\left( {2x} \right) - sin\left( {4x} \right)} \right)} dx\\I &= \frac{1}{2}\left( {\int {sin2x\,dx - \int {sin4x\,dx} } } \right)\end{aligned}\)

We know integration of \(\int {sin\,x}  =  - cos\,x\).

So, \(\int {\sin \,2x = \frac{{ - cos2x}}{2}} \) and \(\int {\sin \,4x = \frac{{ - cos4x}}{4}} \)

\(\begin{aligned}{l}I =& \frac{1}{2}\left( {\frac{{ - cos2x}}{2} - \frac{{cos4x}}{4}} \right) + c\\I &= \frac{{ - cos2x}}{4} - \frac{{cos4x}}{8} + c\end{aligned}\)

Hence, \(I = \int {sin3x\,cosxdx = - \frac{1}{4}cos2x - \frac{1}{8}cos 4x + c} \)