Sin A + Sin B
Sin A + Sin B, an important identity in trigonometry, is used to find the sum of values of sine function for angles A and B. It is one of the sum to product formulas used to represent the sum of sine function for angles A and B into their product form. The result for sin A + sin B is given as 2 sin ½ (A + B) cos ½ (A - B).
Let us understand the sin A + sin B formula and its proof in detail using solved examples.
1. | What is Sin A + Sin B Identity in Trigonometry? |
2. | Sin A + Sin B Sum to Product Formula |
3. | Proof of Sin A + Sin B Formula |
4. | How to Apply Sin A + Sin B? |
5. | FAQs on Sin A + Sin B |
What is SinA + SinB Identity in Trigonometry?
The trigonometric identity sinA + sinB is used to represent the sum of sine of angles A and B, sin A + sin B in the product form using the compound angles (A + B) and (A - B). It says sin A + sin B = 2 sin [(A + B)/2] cWe will study the sin A + sin B formula in detail in the following sections.
Sin A + Sin B Sum to Product Formula
The sin A + sin B sum to product formula in trigonometry for angles A and B is given as,
Sin A + Sin B = 2 sin [½ (A + B)] cos [½ (A - B)]
Here, A and B are angles, and (A + B) and (A - B) are their compound angles.
Proof of SinA + SinB Formula
We can give the proof of sin A + sin B formula (sin A + sin B = 2 sin ½ (A + B) cos ½ (A - B)) using the expansion of sin(A + B) and sin(A - B) formula. We know, using trigonometric identities, ½ [sin(α + β) + sin(α - β)] = sin α cos β, for any angles α and β. From this,
[sin(α + β) + sin(α - β)] = 2 sin α cos β ... (1)
Let us assume that (α + β) = A and (α - β) = B.
⇒ 2α = A + B
⇒ α = (A + B)/2
⇒ 2β = A - B
⇒ β = (A - B)/2
Substituting all these values in (1)
⇒ sinA + sinB = 2 sin ½(A + B) cos ½(A - B)
Hence, proved.
How to Apply Sin A + Sin B?
We can apply the sin A + sin B formula as a sum to the product identity. Let us understand its application using an example of sin 60º + sin 30º. We will solve the value of the given expression by 2 methods, using the formula and by directly applying the values, and compare the results. Have a look at the below-given steps.
- Compare the angles A and B with the given expression, sin 60º + sin 30º. Here, A = 60º, B = 30º.
- Solving using the expansion of the formula sin A + sin B, given as, sin A + sin B = 2 sin ½ (A + B) cos ½ (A - B), we get,
Sin 60º + Sin 30º = 2 sin ½ (60º + 30º) cos ½ (60º - 30º) = 2 sin 45º cos 15º = 2 (1/√2) ((√3 + 1)/2√2) = (√3 + 1)/2. - Also, we know that sin 60º + sin 30º = (√3/2 + 1/2) = (√3 + 1)/2 (from trig table).
Hence, the result is verified.
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Let us have a look at a few examples to understand the concept of sin A + sin B better.
Examples Using Sin A + Sin B Identity
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Example 1: Find the value of sin 200º + sin 20º using sin A + sin B identity.
Solution:
We know, sinA + sinB = 2 sin ½ (A + B) cos ½ (A - B)
Here, A = 200º, B = 20º
sin 200º + sin 20º = 2 sin ½ (200º + 20º) cos ½ (200º - 20º)
= 2 sin 110º cos 90º
= 0 [∵ cos 90º = 0]
Answer: sin 200º + sin 20º = 0
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Example 2: Using the values of angles from the trigonometric table, solve the expression: 2 sin 67.5º cos 22.5º.
Solution:
We can rewrite the given expression as, 2 sin 67.5º cos 22.5º = 2 sin ½ (135)º cos ½ (45)º
Assuming A + B = 135º, A - B = 45º and solving for A and B, we get, A = 90º and B = 45º.
⇒ 2 sin ½ (135)º cos ½ (45)º = 2 sin ½ (90º + 45º) cos ½ (90º - 45º)
We know, sin A + sin B = 2 sin ½ (A + B) cos ½ (A - B)
2 sin ½ (90º + 45º) cos ½ (90º - 45º) = sin 90º + sin 45º = 1 + (1/√2).
Answer: 2 sin 67.5º cos 22.5º = 1 + (1/√2)
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Example 3: Prove that [(sin A - sin B)/(cos A + cos B)] + [(cos A - cos B)/(sin A + sin B)] = 0.
Solution:
Here, L.H.S. = [(sin A - sin B)/(cos A + cos B)] + [(cos A - cos B)/(sinA + sinB)]
= [2 cos ½ (A + B) sin ½ (A - B)]/[2 cos ½ (A + B) cos ½ (A - B)] + [- 2 sin ½ (A + B) sin ½ (A - B)]/[2 sin ½ (A + B) cos ½ (A - B)]
= [sin ½ (A - B)]/[cos ½ (A - B)] + [- sin ½ (A - B)]/[cos ½ (A - B)]
= [sin ½ (A - B)]/[cos ½ (A - B)] - [ sin ½ (A - B)]/[cos ½ (A - B)]
= 0
= R.H.S.
Hence, proved.
Answer: The given identity is proved.
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Example 4: Verify the given expression using expansion of sin A + sin B: sin 70º + cos 70º = 2 sin 45º cos 25º
Solution:
We have, L.H.S. = sin 70º + cos 70º
Since, cos 70º = cos(90º - 20º) = sin 20º
⇒ sin 70º + cos 70º = sin 70º + sin 20º
Using sin A + sin B = 2 sin ½ (A + B) cos ½ (A - B)
⇒ sin 70º + sin 20º = 2 sin ½ (70º + 20º) cos ½ (70º - 20º)
= 2 sin 45º cos 25º
= R.H.S.
Hence, verified.
Answer: The given equation is proved.
FAQs on Sin A + Sin B
What is the Value of Sin A Plus Sin B?
Sin A plus Sin B is an identity or trigonometric formula, used in representing the sum of sine of angles A and B, Sin A + Sin B in the product form using the compound angles (A + B) and (A - B). Here, A and B are angles.
What is the Formula of SinA + SinB?
SinA + SinB formula, for two angles A and B, can be given as sinA + sinB = 2 sin ½ (A + B) cos ½ (A - B). Here, (A + B) and (A - B) are compound angles.
What is the Product Form of Sin A + Sin B in Trigonometry?
The product form of sin A + sin b formula is given as, sin A + sin B = 2 sin ½ (A + B) cos ½ (A - B), where A and B are any given angles.
How to Prove the Expansion of SinA + SinB Formula?
The expansion of sin A + sin B, given as sinA + sinB = 2 sin ½ (A + B) cos ½ (A - B), can be proved using the 2 sin α cos β product identity in trigonometry. Click here to check the detailed proof of the formula.
How to Use Sin A + Sin B Formula?
To use sin A + sin B identity in a given expression, compare the sin a + sin b formula, sin A + sin B = 2 sin ½ (A + B) cos ½ (A - B) with given expression and substitute the values of angles A and B.
What is the Application of SinA + SinB Formula?
SinA + SinB formula can be applied to represent the sum of sine of angles A and B in the product form of sine of (A + B) and cosine of (A - B), using the formula, sin A + sin B = 2 sin ½ (A + B) cos ½ (A - B).
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