Product to Sum Formulae - JEE Important Topic

VSAT 2022

What is The Product to Sum Formula?

The product to sum formulas are the trigonometric identities that are used to expand the product of cosine and sine functions as a sum of cosine and sine functions or vice versa. These are used to solve some of the fundamental or critical trigonometric functions. The product to sum or (POS) formulas are derived from sum and difference formulas of trigonometry. 


The sum and difference formulas in trigonometry are quite similar except for their signs. The sum to product or SOP is another way to simplify the complicated trigonometric functions. The sum of product or SOP are derived from product to sum formulas. Therefore, just like product to sum formulas we can also express trigonometric functions as sum to products of cosine and sine functions. Let us further look into these concepts. 


What is The Product To Sum?

As the name suggests, the product to sum is nothing but expressing the product of sine and cosine functions as a sum. It is a concept that is used to simplify trigonometric functions as sometimes we may need to express the trigonometric functions a sum. The product to sum can also be sometimes called a POS formula.


The fundamental product to sum identities are as given below:

$\begin{align} &\sin (A) \cos (B)=\dfrac{1}{2}(\sin (A+B)+\sin (A-B)) \\ &\cos (A) \sin (B)=\dfrac{1}{2}(\sin (A+B)-\sin (A-B)) \\ &\sin (A) \sin (B)=\dfrac{1}{2}(\cos (A-B)+\cos (A+B)) \\ &\cos (A) \cos (B)=\dfrac{1}{2}(\cos (A+B)-\cos (A-B)) \end{align}$


Product To Sum Examples

1. Find the value of sin 45o sin 15o using the product to sum formulas.

Solution:

Using the third product to sum formulas above we have,

$\sin (A) \sin (B)=\dfrac{1}{2}(\cos (A-B)+\cos (A+B))$

Substituting the values we get,

$\sin 45^{\circ} \sin 15^{\circ}=\dfrac{1}{2}\left(\cos 30^{\circ}-\cos 60^{\circ}\right)$

0.866+0.5=1.366

 

2. Write the product of the $2 \cos \left(\dfrac{7 x}{2}\right) \cos \left(\dfrac{3 x}{2}\right)$ as a sum

Solution:

Using the 4th identity of the product to sum from above, we get

$\begin{align} &\cos (A) \cos (B)=\dfrac{1}{2}(\cos (A+B)-\cos (A-B)) \\ &=\frac{1}{2} \times 2(\cos 5 x-\cos 2 x) \\ &=(\cos 5 x-\cos 2 x) \end{align}$

 

State The Product To Sum Formula and their Derivation

To derive the product to sum derivations we need to know the sum and difference trigonometric identities. These are as listed below:

$\begin{align} &\sin (A+B)=\sin (A) \cos (B)+\cos (A) \sin (B) \ldots \ldots(a) \\ &\cos (A+B)=\cos (A) \cos (B)-\sin (A) \sin (B) \ldots \ldots(b) \\ &\sin (A-B)=\sin (A) \cos (B)-\cos (A) \sin (B) \ldots \ldots(c) \\ &\cos (A-B)=\cos (A) \cos (B)+\sin (A) \sin (B) \ldots \ldots(d) \end{align}$


(a) Derivation for $\sin (A) \cos (B)=\dfrac{1}{2} \sin (A+B)+\sin (A-B)$

Adding the equations (a) and (c), we get

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

Divide both sides by 2, and we get

$\sin (A) \cos (B)=\dfrac{1}{2} \sin (A+B)+\sin (A-B)$

 

(b) Derivation for $\cos (A) \sin (B)=\frac{1}{2} \sin (A+B)-\sin (A-B)$

Subtracting (c) from (a), we get

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

Divide both sides by 2, and we get

$\cos (A) \sin (B)=\dfrac{1}{2} \sin (A+B)-\sin (A-B)$

 

(c) Derivation for$\cos (A) \cos (B)=\dfrac{1}{2} \cos (A+B)+\cos (A-B)$

Adding (b) and (d), we get $\cos (A+B)+\cos (A-B)=2 \cos (A) \cos (B)$

Divide both sides by 2, and we get

$\cos (A) \cos (B)=\dfrac{1}{2} \cos (A+B)+\cos (A-B)$

 

(d) Derivation for $\sin (A) \sin (B)=\dfrac{1}{2} \cos (A-B)-\cos (A+B)$

Subtracting (b) from (d), $\cos (A-B)-\cos (A+B)=2 \sin (A) \sin (B)$

Divide both sides by 2, and we get

$\sin (A) \sin (B)=\dfrac{1}{2} \cos (A-B)-\cos (A+B)$


What is the Sum of Product?

As the name suggests, the sum of product is used to express the sum of cosine and sine functions as a product. These are used to simplify critical trigonometric functions. The fundamental sum to product formulas are as listed below:

$\begin{align} &\sin (A)+\sin (B)=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right) \\ &\sin (A)-\sin (B)=2 \sin \left(\dfrac{A-B}{2}\right) \cos \left(\dfrac{A+B}{2}\right) \\ &\cos (A)-\cos (B)=-2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right) \\ &\cos (A)+\cos (B)=2 \cos \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right) \end{align}$


Sum of Product Examples

1. Evaluate the value of sin 15° + sin 45° using the sum to product formula.

Solution: We can find the value of sin 15° + sin 45° by using the first identity of the sum to produce the identity listed above. 

We know that $\sin (A)+\sin (B)=2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)$

Substituting the values 15° and 45° above, we have

$\begin{align} &\sin 15^{\circ}+\sin 75^{\circ}=2 \sin \left(\dfrac{90^{\circ}}{2}\right) \cos \left(\dfrac{60^{\circ}}{2}\right) \\ &=2 \sin 45^{\circ} \cos 30^{\circ} \end{align}$  

Now we know the identity cos(-x) = cos x, therefore cos(- 30°)= cos 30° 

Putting the values of 45° and 15° from the standard trigonometric table we get

$\begin{align} &=2 \times 1 / \sqrt{2} \times \sqrt{3} / 2 \\ &2 \times \dfrac{\sqrt{3}}{2} \times \dfrac{1}{2}=\dfrac{\sqrt{3}}{2} \end{align}$

Therefore the answer for $\sin 15^{\circ}+\sin 45=\dfrac{\sqrt{3}}{2}$

 

2. Express cos 3x - cos x as a product of trigonometric function using the SOP formulas. 

Solution: We know from the third SOP identity listed above that

$\cos (A)-\cos (B)=-2 \sin \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)$

Substituting the values 3x and x in the formula we get, 

$\cos 3 x-\cos x=-2 \sin \dfrac{3 x+x}{2} \sin \dfrac{(3 x-x)}{2}=-2 \sin ^{2} x$

Therefore, cos 3x - cos x = -2 sin2x

 

3. Use sum to product or SOP formula to express the sum of cos 4x + cos 2x as a product of the same.
Solution:

To express the above equation cos (4x) + cos(2x) we use the identity 

$\cos (A)+\cos (B)=2 \cos \left(\dfrac{A+B}{2}\right) \cos \left(\dfrac{A-B}{2}\right)$

Assuming A=4x and B=2x in the identity above, we get
cos 8x + cos 2x =
$2 \cos \dfrac{(4 x+2 x)}{2} \cos \dfrac{(4 x-2 x)}{2}$

$2 \cos \dfrac{6 x}{2} \cos \dfrac{2 x}{2}$

$2 \cos 3 x \cos x$

Thus, the sum cos (4x) + cos(2x) = $2 \cos 3 x \cos x$

Conclusion

The article presents a thorough explanation of the different formulas or identities to solve some of the complex trigonometric functions. Suppose if we are to solve for a product of sin 75 and sin 15 degree, it would be difficult to solve these values individually as these are not the standard trigonometric angles. Additionally, we can derive the sum of product formulas for the product to sum formulas in trigonometry. Thus we can make use of the product to sum or sum to product identities to express these as a sum. It makes the calculations much easier and daves a lot of time.

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FAQs on Product to Sum Formulae - JEE Important Topic

1. State the application of product to sum formulas?

As we learned from above, the product to sum formulas are useful in expressing the product of cosine and sine functions as a sum. These are quick and useful identities that are used to make integration easier. Integration as we know can be simplified if we integrate a sum rather than a product of trigonometric identities.  These formulas are hence useful in integration as integrating a sum is pretty easier when compared to integrating a product.

2. What is the relation between difference and sum of angles in trigonometry?

The difference and sum values in trigonometry are used to find the values of angles. These could be the values of any angles that are not a part of the standard trigonometric table i.e 0, 30, 45,60, and 90 degrees. The practical use of this is to find the exact values of trigonometric functions that are expressed as the sum or difference of sin and cosine functions. Further these can also be used to derive products to sum formulas. The sum of product formulas can however be derived from the product to sum formulas.

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