Product To Sum Formulas

What are Product-to-sum/difference formulae?

The Product formula is used to express the Product of sine and cosine functions into the sum or difference of sine and cosine functions. The sum and difference formulas of sine and cosine functions are added or subtracted to derive these identities. The product-to-sum identities can be used to simplify the trigonometric expression.

Product into Sum/Difference Formulas

Product-to-sum formulas provide a powerful tool for simplifying trigonometric expressions involving products of sines and cosines, and the product to sum formulas are:

1. $2 \cos \alpha \cos \beta=[\cos (\alpha-\beta)+\cos (\alpha+\beta)]$
2. $2 \sin \alpha \sin \beta=[\cos (\alpha-\beta)-\cos (\alpha+\beta)]$
3. $2 \sin \alpha \cos \beta=[\sin (\alpha+\beta)+\sin (\alpha-\beta)]$
4. $2 \cos \alpha \sin \beta=[\sin (\alpha+\beta)-\sin (\alpha-\beta)]$

where, $\alpha$ and $\beta$ are two angles of a triangle
1) $2 \cos \alpha \cos \beta=\cos (\alpha-\beta)+\cos (\alpha+\beta)$

This formula involves the conversion of the product of cosine functions of two different angles into a sum of the cosine angle.
2) $2 \sin \alpha \sin \beta=\cos \left(\alpha-\beta\right)-\cos \left(\alpha+\beta\right)$

This formula involves the conversion of the product of sine functions of two different angles into a difference in the cosine angle.
3) $2 \sin \alpha \cos \beta=\sin (\alpha+\beta)+\sin (\alpha-\beta)$

This formula involves the conversion of the product of sine and cosine functions of two different angles into a sum of the sine angle.
4) $2 \cos \alpha \sin \beta=\sin \left(\alpha{+} \beta\right)-\sin (\alpha-\beta)$

This formula involves the conversion of the product of sine and cosine functions of two different angles into a different of sine angle.

Proof of Product-to-sum formulae

sin (A + B) = sin A cos B + cos A sin B ... (1)
sin (A - B) = sin A cos B - cos A sin B ... (2)
cos (A + B) = cos A cos B - sin A sin B ... (3)
cos (A - B) = cos A cos B + sin A sin B ... (4)

1. Deriving the formula sin A cos B = (1/2) [ sin (A + B) + sin (A - B) ]

Adding the equations (1) and (2), we get

sin (A + B) + sin (A - B) = 2 sin A cos B

Dividing both sides by 2,

sin A cos B = (1/2) [ sin (A + B) + sin (A - B) ]

2. Deriving the formula cos A sin B = (1/2) [ sin (A + B) - sin (A - B) ]

Subtracting (2) from (1),

sin (A + B) - sin (A - B) = 2 cos A sin B

Dividing both sides by 2,

cos A sin B = (1/2) [ sin (A + B) - sin (A - B) ]

3. Deriving the formula cos A cos B = (1/2) [ cos (A + B) + cos (A - B) ]

Adding the equations (3) and (4), we get

cos (A + B) + cos (A - B) = 2 cos A cos B

Dividing both sides by 2,

cos A cos B = (1/2) [ cos (A + B) + cos (A - B) ]

4. Deriving the formula sin A sin B = (1/2) [ cos (A - B) - cos (A + B) ]

Subtracting (3) from (4),

cos (A - B) - cos (A + B) = 2 sin A sin B

Dividing both sides by 2,

sin A sin B = (1/2) [ cos (A - B) - cos (A + B) ]

Product of cosines

$
\begin{array}{r}
\cos \alpha \cos \beta+\sin \alpha \sin \beta=\cos (\alpha-\beta) \\
+\cos \alpha \cos \beta-\sin \alpha \sin \beta=\cos (\alpha+\beta) \\
\hline 2 \cos \alpha \cos \beta=\cos (\alpha-\beta)+\cos (\alpha+\beta)
\end{array}
$

Product of sine and cosine

$
\begin{gathered}
\sin (\alpha+\beta)=\sin (\alpha) \cos (\beta)+\cos (\alpha) \sin (\beta) \\
+\sin (\alpha-\beta)=\sin (\alpha) \cos (\beta)-\cos (\alpha) \sin (\beta) \\
\hline \sin (\alpha+\beta)+\sin (\alpha-\beta)=2 \sin (\alpha) \cos (\beta)
\end{gathered}
$

Product of cosine

$
\begin{gathered}
\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta \\
-\quad \cos (\alpha+\beta)=-(\cos \alpha \cos \beta-\sin \alpha \sin \beta) \\
\hline \cos (\alpha-\beta)-\cos (\alpha+\beta)=2 \sin \alpha \sin \beta
\end{gathered}
$

Recommended Video Based on Product to Sum/Difference

Solved Example Based on Product to Sum/Difference

Example 1: The value of $
\cos \left(\frac{2 \pi}{7}\right)+\cos \left(\frac{4 \pi}{7}\right)+\cos \left(\frac{6 \pi}{7}\right)
$ is equal to? [JEE MAINS 2022]
Solution:

Using Summation of cosine series

$
\begin{aligned}
\cos \frac{2 \pi}{7}+\cos \frac{4 \pi}{7}+\cos \frac{6 \pi}{7} & =\frac{\sin 3 \times \frac{2 \pi}{2 \times 7}}{\sin \frac{2 \pi}{2 \times 7}} \times \cos \left(\frac{\frac{2 \pi}{7}+\frac{6 \pi}{7}}{2}\right)
\end{aligned} $

$\begin{aligned} & =\frac{\sin \frac{5 \pi}{7}}{\sin \frac{\pi}{7}} \times \cos \frac{4 \pi}{7} \\ & =\frac{\sin \left(\pi-\frac{4 \pi}{7}\right) \cos \frac{4 \pi}{7}}{\sin \frac{\pi}{7}}\end{aligned}$
$
\begin{aligned}
& =\frac{2 \sin \frac{4 \pi}{7} \cos \frac{4 \pi}{7}}{2 \sin \frac{\pi}{7}} \\
& =\frac{\sin \frac{8 \pi}{7}}{2 \sin \frac{\pi}{7}}=-\frac{1}{2}
\end{aligned}
$
Hence the answer is $1 / 2$

Example 2: The value of $\cos ^2 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^2 50^{\circ}$ is [JEE MAINS 2019]
Solution: $\cos ^2 10^{\circ}-\cos 10^{\circ} \cos 50^{\circ}+\cos ^2 50^{\circ}$

$
\begin{aligned}
& \Rightarrow \frac{1+\cos 20^{\circ}}{2}+\frac{1+\cos 100^{\circ}}{2}-\cos 10^{\circ} \cos 50^{\circ} \\
& \Rightarrow \frac{1}{2}\left[2+\cos 20^{\circ}+\cos 100^{\circ}-2 \cos 10^{\circ} \cos 50^{\circ}\right] \\
& \Rightarrow \frac{1}{2}\left[2+\cos 100^{\circ}+\cos 20^{\circ}-\cos 60^{\circ}-\cos 40^{\circ}\right] \\
& \Rightarrow \frac{1}{2}\left[\frac{3}{2}+2 \cos 60^{\circ} \cdot \cos 40^{\circ}-\cos 40^{\circ}\right] \\
& \Rightarrow \frac{1}{2}\left[\frac{3}{2}+2 \times \frac{1}{2} \cos 40^{\circ}-\cos 40^{\circ}\right]=\frac{3}{4}
\end{aligned}
$
Hence, the answer is $3 / 4$

Example 3: If $x+\frac{1}{x}=2 \cos \theta$, then $x^3+\frac{1}{x^3}$ ?
Solution

$
\begin{aligned}
&\left(x+\frac{1}{x}\right)^3=x^3+\frac{1}{x^3}+3 x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right) \quad \because(x+y)^3=x^3+y^3+3 x y(x+y) \\
& x^3+\frac{1}{x^3}=\left(x+\frac{1}{x}\right)^3-3 x \cdot \frac{1}{x}\left(x+\frac{1}{x}\right) \quad \text { Given }\left(x+\frac{1}{x}\right)=2 \cos \theta \\
& x^3+\frac{1}{x^3}=(2 \cos \theta)^3-3 \cdot 2 \cos \theta \\
&=2\left[4 \cos ^3 \theta-3 \cos \theta\right] \\
&=2 \cos 3 \theta
\end{aligned}
$
Hence, the answer is $2 \cos 3 \theta$

Example 4: The value of $\sin \left(18^{\circ}\right)+\sin \left(72^{\circ}\right)-\sqrt{2} \cos \left(27^{\circ}\right)$ is
Solution

$
\begin{aligned}
& \sin \left(18^{\circ}\right)+\sin \left(72^{\circ}\right)-\sqrt{2} \cos \left(27^{\circ}\right) \\
& =2 \sin \left(45^{\circ}\right) \cos 27^{\circ}-\sqrt{2} \cos \left(27^{\circ}\right) \\
& =0
\end{aligned}
$
Hence, the answer is 0.

Example 5: if $\sin (3 x)+\sin (2 x)-\sin (x)=0$, then find the number of solutions in $[0, \pi]$ ?
Solution

$
\Rightarrow \sin (3 x)+\sin (2 x)-\sin (x)=0
$

$
\begin{aligned}
& \Rightarrow \sin (3 x)-\sin (x)+\sin (2 x)=0 \\
& \Rightarrow 2 \sin (x) \cos (2 x)+\sin (2 x)=0 \\
& \Rightarrow 2 \sin (x) \cos (2 x)+2 \sin (x) \cos (x)=0 \\
& \Rightarrow 2 \sin (x)\{\cos (2 x)+\cos (x)=0 \\
& \Rightarrow 2 \sin (x)\left\{2 \cos \left(\frac{3 x}{2}\right) \cos \left(\frac{x}{2}\right)\right\}=0 \\
& \Rightarrow x=0, \frac{\pi}{3}, \pi
\end{aligned}
$

Hence, the answer is the 3.