TRANSFORMATION FORMULAE

Formula to transform the product into sum or difference

The sine formula of the sum and difference of two angles is written as follows:

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Add both the expression as follows:

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Then, subtract expression (2) from expression (1) as follows:

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Similarly consider the cosine formula of the sum and difference of two angles is written as follows:

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Add both the expression as follows:

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Then, subtract expression (3) from expression(4) as follows:

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Therefore, following formula is obtained to transform the product into sum or difference

1. 2\sin \alpha \cos \beta  = \sin \left( {\alpha  + \beta } \right) + \sin \left( {\alpha  - \beta } \right)

2. 2\cos \alpha \sin \beta  = \sin \left( {\alpha  + \beta } \right) - \sin \left( {\alpha  - \beta } \right)

3. 2\cos \alpha \cos \beta  = \cos \left( {\alpha  + \beta } \right) + \cos \left( {\alpha  - \beta } \right)

4. 2\sin \alpha \sin \beta  = \cos \left( {\alpha  - \beta } \right) - \cos \left( {\alpha  + \beta } \right)

Formula to transform the sum or difference into product

Take the formula

2\sin \alpha \cos \beta  = \sin \left( {\alpha  + \beta } \right) + \sin \left( {\alpha  - \beta } \right)

Assume\alpha  + \beta  = C{\rm{ and }}\alpha  - \beta  = D,

Then,

\alpha  = \frac{{C + D}}{2}{\rm{ and }}\beta  = \frac{{C - D}}{2}

Substitute the values in the formulae obtained in transforming the product into sum or difference

1. 2\sin \left( {\frac{{C + D}}{2}} \right)\cos \left( {\frac{{C - D}}{2}} \right) = \sin C + \sin D

2. 2\cos \left( {\frac{{C + D}}{2}} \right)\sin \left( {\frac{{C - D}}{2}} \right) = \sin C - \sin D

3. 2\cos \left( {\frac{{C + D}}{2}} \right)\cos \left( {\frac{{C - D}}{2}} \right) = \cos C + \cos D

4. 2\sin \left( {\frac{{C + D}}{2}} \right)\sin \left( {\frac{{C - D}}{2}} \right) = \cos C - \cos D

Question 1

Write down the difference of \sin 60^\circ  - \sin 40^\circ as a product

Solution

Use the formula

2\cos \left( {\frac{{C + D}}{2}} \right)\sin \left( {\frac{{C - D}}{2}} \right) = \sin C - \sin D

Here, C = 60^\circ {\rm{ and }}D = 40^\circ substitute the value in the formula as follows:

\begin{array}{l}
2\cos \left( {\frac{{60 + 40}}{2}} \right)\sin \left( {\frac{{60 - 40}}{2}} \right) = \sin 60 - \sin 40\\
2\cos \left( {\frac{{100}}{2}} \right)\sin \left( {\frac{{20}}{2}} \right) = \sin 60 - \sin 40\\
2\cos 50 \cdot \sin 10 = \sin 60 - \sin 40
\end{array}

Question 2

Evaluate \sin 120^\circ \sin 20^\circ  using an appropriate product sum identity

Solution

Use the formula

2\sin \alpha \sin \beta  = \cos \left( {\alpha  - \beta } \right) - \cos \left( {\alpha  + \beta } \right)

Rewritten as follows:

\sin \alpha \sin \beta  = \frac{1}{2}\left[ {\cos \left( {\alpha  - \beta } \right) - \cos \left( {\alpha  + \beta } \right)} \right]

Here\alpha  = 120^\circ {\rm{ and }}\beta  = 20^\circ , substitute the value in the formula as follows:

\begin{array}{c}
\sin 120^\circ \sin 20^\circ  = \frac{1}{2}\left[ {\cos \left( {120^\circ  - 20^\circ } \right) - \cos \left( {120^\circ  + 20^\circ } \right)} \right]\\
 = \frac{1}{2}\left[ {\cos \left( {100^\circ } \right) - \cos \left( {140^\circ } \right)} \right]\\
 = \frac{1}{2}\left[ { - {\rm{0}}.{\rm{1736 + 0}}{\rm{.7660}}} \right]\\
 = \frac{1}{2}\left[ {0.5924} \right]\\
\sin 120^\circ \sin 20^\circ  = 0.2962
\end{array}

Post Author: E-Maths