Sin A Minus B or sin (A – B) is one of the many common compound identities. It is used in to find the sin of difference of any two angles as difference of product of sine and cosine of individual angles. In this article, we will discuss this formula in detail, including proof and solved examples.
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What is Sin(a-b)?
The sine of the difference of two angles a and b, denoted as sin (a − b). It is a compound angle formula where we want to denote the sine value of the difference between two angles. Sin (a-b) is also called the difference formula. Sin a minus b is equal to the product of sin a and cos b minus the product of cos a and sin b.
Formula for Sin (a-b)
This formula provides us with the relationship between the sine value of two angles and their cosine values.
sin (a-b) = sin(a) . cos(b) – cos(a) . sin(b)
This formula allows us to express the sine of the difference of two angles in terms of the sines and cosines of the individual angles.
Proof of Sin(a – b) Formula
To prove: sin (a – b) = sin a cos b – cos a sin b
Construction: Let OX be a rotating line. Rotate it about O in the anti-clockwise direction to form the rays OY and OZ such that ∠XOZ = a and ∠YOZ = b. Then ∠XOY = a – b.
Take a point P on the ray OY, and draw perpendiculars PQ and PR to OX and OZ respectively. Again, draw perpendiculars RS and RT from R upon OX and PQ respectively.
Proof: We will see how we have written ∠TPR = a in the above figure.
From the right triangle OPQ, ∠OPQ = 180 – (90 + a – b) = 90 – a + b;
From the right triangle OPR, ∠OPR = 180 – (90 + b) = 90 – b
Now, from the figure, ∠OPQ, ∠OPR, and ∠TPR are the angles at a point on a straight line and hence they add up to 180 degrees.
∠OPQ + ∠OPR + ∠TPR = 180
⇒ (90 – a + b) + (90 – b) + ∠TPR = 180
⇒ 180 – a + ∠TPR = 180
⇒ ∠TPR = a
In triangle ORS, we get
sin a = RS/OR
In triangle TPR, we get
cos a = TP/PR
In triangle ORP, we get
cos b = OR/OP
In triangle OPR, we get
sin b = PR/OP
Now, from the right-angled triangle PQO we get,
sin (a – b) = PQ/OP
⇒ sin (a – b) = (QT-TP)/OP
⇒ sin (a – b) = QT/OP – TP/OP
⇒ sin (a – b) = RS/OP – TP/OP
⇒ sin (a – b) = RS/OR ∙ OR/OP – TP/PR ∙ PR/OP
⇒ sin (a – b) = sin a cos b – cos a sin b
Therefore, sin (a – b) = sin a cos b – cos a sin b.
Alternate Method
We are going to use the sin a plus b formula to prove this particular sine formula.
As we know, identity for sin (x + y) is given as
sin(x + y) = sin(x) . cos(y) + cos(x) . sin(y)
Now let’s substitute x = a and y = -b, we will get:
sin(a – b) = sin(a) cos(−b) + cos(a) sin(−b)
As cos(-b) = cos (b) and sin(-b) = -sin (b)
sin(a – b) = sin(a) cos(b) – cos(a) sin(b)
Which is the required identity.
How to Apply Sin(a – b)?
If we want to apply the sin(a-b) formula we must follow up these steps:
Step 1: Identify the values for angle a and b
Step 2: Substitute the values into sin(a – b) formula
Step 3: Calculate the sine values of the angle a and b using the trigonometric tables.
Step 4: Evaluate the expression to get the result.
Let’s consider an example for better understanding.
Example: Verify sin(60° – 30°) = sin 30°.
Solution:
Compare the sin(a – b) expression with the given expression to identify the angles ‘a’ and ‘b’. Here, a = 60º and b = 30º.
We know, sin (a – b) = sin a cos b – cos a sin b.
⇒ sin(60° – 30°) = sin 60° cos° – sin 30° cos 60°
Since, sin 30° = 1/2, sin 60° = √3/2, cos 30° = √3/2, cos 60° = 1/2
⇒ sin(60° – 30°) = (√3/2)(√3/2) – (1/2)(1/2) = 3/4 – 1/4 = 2/4 = 1/2
Also, we know that sin(60° – 30°) = sin 30° = 1/2.
Some Other Similar Identities
Some other similar identities includes:
- sin (A – B) = sin A cos B – cos A sin B
- cos (A + B) = cos A cos B – sin A sin B
- cos (A – B) = cos A cos B + sin A sin B
- tan (A + B) = (tan A + tan B)/(1 – tan A tan B)
- tan (A – B) = (tan A – tan B)/(1 + tan A tan B)
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Examples Using Sin(a – b) Formula
Example 1: Find the value of Sin 15°.
Solution:
To find the value of sin(15°) using the sin(a – b) identity, we can express 15° as the sum of two common angles. One possible representation is 45° – 30°.
The sin(a – b) identity is:
sin (a – b) = sin (a) cos(b) – cos(a)sin (b)
Substituting a = 45° and b = 30°, we get:
sin (45° – 30°) = sin (45°) cos(30°) – cos(45°)sin (30°)
We know that sin (45°) = cos(45°) = 1/√2 and cos(30°) = √3/2 and sin (30°) = 1/2.
So, substituting these values, we have:
⇒ sin (15°) = 1/√2 × √3/2 – 1/√2 × 1/2
⇒ sin (15°) = √3/2√2 – 1/2√2
⇒ sin (15°) = (√3 – 1)/2√2 = (√3 – 1)√2/4
Thus, using the sin(a – b) identity, we find that sin(15°) = (√3 – 1)/2√2 or (√3 – 1)√2/4
Example 2: Determine the value of Sin 30° using the Sin a minus b formula
Solution:
We can write sin 30 as Sin(90° – 60°)
Using sin a minus b formula for Sin(90° – 60°), we get
sin 90.cos60 – cos90.sin60
1.1/2 – 0.√3/2 = 1/2 – 0
Practice Questions on Sin (a – b)
Q1: Determine sin (45°-30°)
Q2: Calculate sin (5π/3 – π/4)
Q3: Find sin (120°- 60°)
Q4: Compute sin (45°) using Sin (a – b) Formula.
FAQs on Sin A minus B Formula
What is the formula for sin (A – B) ?
The formula for sin (A – B) is given by:
sin (A – B) = sin (A)cos (B) – cos (A)sin (B)
What does A and B represent in the formula?
A and B represent angles. The formula calculates the sine of the difference of two angles A and B .
How is the sin (A – B) formula derived?
The formula is derived from the angle addition formula for sine, which states that sin (A + B) = sin (A)cos (B) + cos (A)sin (B) . By replacing B with -B , we get sin (A – B) .
What is the geometric interpretation of sin (A – B) ?
Geometrically, sin (A – B) represents the vertical distance between two points on the unit circle corresponding to angles A and B, with A being greater than B.