2cosAsinB is one of the important trigonometric formulas which is equal to sin (A + B) - sin (A-B). It is a fundamental identity in trigonometry, often used in simplifying expressions and soving the equations. 2cosAsinB Formula is used in trigonometry to solve various equations and problems more efficiently. This formula is found using the sum and difference of the sine function.
In this article, we have covered, in brief, trigonometric ratios, 2Cos(a)Sin(b) Formula, its derivation, and others in detail.
Table of Content
Trigonometric Ratios
Trigonometric ratios are ratios of sides in a triangle and there are six trigonometric ratios. In a right-angle triangle, the six trigonometric ratios are defined as:
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- sin θ = (Opposite Side/Hypotenuse = AB/AC
- cos θ = Adjacent Side/Hypotenuse = BC/AC
- tan θ = Opposite side/adjacent side = AB/BC
- cosec θ = 1/sin θ = Hypotenuse/Opposite Side = AC/AB
- sec θ = 1/cos θ = Hypotenuse/Adjacent Side = AC/BC
- cot θ = 1/tan θ = Adjacent Side/Opposite Side = BC/AB
2Cos(A)Sin(B) Formula
2cosAsinB formula is a trigonometric formula that is used to simplify trigonometric expressions and also solve complex integrals and derivatives of trigonometric expressions. The 2cosAsinB formula is equal to the difference between the angle sum and the angle difference of the sine functions, i.e., for two angles A and B,
Sin(b)-Formula.webp)
2cosasinb formula is,
2 cos A.sin B = sin (A + B) - sin (A - B)
From the formula, we can observe that twice the product of a cosine function and a sine function is converted into the difference between the angle sum and the angle difference of the sine functions. With the help of the 2 cos A sin B formula, we can extract the formula of cos A sin B.
cos A.sin B = ½ [sin (A + B) - sin (A - B)]
Derivation of 2CosaSinb Formula
We can derive the 2cosasinb formula with the help of the sum and difference of formulae of the sine function.
- sin (A + B) = sin A cos B + cos A sin B... (1)
- sin (A - B) = sin A cos B - cos A sin B...(2)
Now subtract the equation (2) from the equation (1)
⇒ sin (A + B) - sin (A - B) = (sin A cos B + cos A sin B) - (sin A cos B - cos A sin B)
⇒ sin (A + B) - sin (A - B) = sin A cos B + cos A sin B - sin A cos B + cos A sin B
⇒ sin (A + B) - sin (A - B) = cos A sin B + cos A sin B
⇒ sin (A + B) - sin (A - B) = 2 cos A sin B
Hence, 2 cos A sin B = sin (A + B) - sin (A - B)
Articles Related to 2cosAsinB Formula:
Examples on 2CosaSinb Formula
Example 1: Solve the integral of 2cos 3x sin (5x/2).
Solution:
Integral of 2cos 3x sin (5x/2) = ∫2 cos 3x sin (5x/2) dx
From 2cosasinb formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
2 cos 3x sin (5x/2) = sin (3x + (5x/2)) - sin (3x - (5x/2))
= sin (11x/2) - sin (x/2)
Now, ∫2 cos 3x sin (5x/2)) dx = ∫[sin (11x/2) - sin (x/2)] dx
= ∫sin (11x/2) dx - ∫sin (x/2) dx
= -2/11 cos (11x/2) - (-2 cos (x/2)) {∫sin (ax) = -1/a cos (ax) + c}
= 2[cos (x/2) - 1/11 cos (11x/2)]
Hence, the integral of 2 cos 3x sin (5x/2) = 2[cos (x/2) - 1/11 cos (11x/2)]
Example 2: Express 5cos (7x/2) sin 3x in terms of the sine function.
Solution:
From 2cosasinb formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
Now, 5 cos (7x/2) sin 3x = 5/2 [2 cos (7x/2) sin 3x]
= 5/2 [sin (7x/2 + 3x) - sin (7x/2 - 3x)]
= 5/2 [sin (13x/2) - sin (x/2)]
Hence, 5 cos (7x/2) sin 3x = 5/2 [sin (13x/2) - sin (x/2)].
Example 3: Find the value of the expression 4 cos (27.5°) sin (62.5°) using the 2cosasinb formula.
Solution:
4 cos (27.5°) sin (62.5°) = 2 [2 cos (27.5°) sin (62.5°)]
From 2cosasinb formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
Now, 2 [2 cos (27.5°) sin (62.5°)] = 2 [sin (27.5° + 62.5°) - sin (27.5° - 62.5°)]
=2 [sin (90°) - sin (-35°)]
= 2 [sin 90°+ sin 35°] {Since, sin (-θ) = - sin θ}
= 2 [1 + 0.5735] {Since, sin 35° = 0.5735, sin 90° = 1}
= 3.147
Hence, 4 cos (27.5°) sin (62.5°) = 3.147
Example 4: Find the derivative of 7 cos 4x sin 11x.
Solution:
Derivative of 7 cos 4x sin 11x = d(7 cos 4x sin 11x)/dx
From 2cosasinb formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
Now, 7 cos 4x sin 11x = 7/2 [2 cos 4x sin 11x ]
= 7/2 [sin (4x + 11x) - sin (4x - 11x)]
= 5/2 [sin (15x) - sin (-7x)]
= 5/2 [sin (15x) + sin (7x)] {Since, sin (-θ) = - sin θ}
Now, d(7 cos 4x sin 11x)/dx = d{5/2 [sin 15x + sin 7x]}/dx
= 5/2{d(sin 15x)/dx + d( sin 7x)/dx}
= 5/2 [15 cos 15x + 7 cos 7x] {Since, d(sin ax)/dx = a cos ax}
= 37.5 cos 15x + 17.5 cos 7x
Hence, the derivative of 7 cos 4x sin 11x = [37.5 cos 15x + 17.5 cos 7x] .
Example 5: Express 2 cos 14x sin (3x/2) in terms of the sine function.
Solution:
From 2cosasinb formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
Now, 2 cos 14x sin (3x/2) = sin (14x + 3x/2) - sin (14x - 3x/2)
= sin [(28x + 3x)/2] - sin [(28x - 3x)/2]
= sin (31x/2) - sin (25x/2)
Hence, 2 cos 14x sin (3x/2) = [sin (31x/2) - sin (25x/2)].
Example 6: Solve 6 sin (52.5 °) sin (127.5°) using the 2cosasinb formula.
Solution:
6 sin (52.5 °) sin (127.5°) = 3 [2 sin (52.5 °) sin (127.5°)]
From 2cosasinb formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
Now, 3 [2 sin (52.5 °) sin (127.5°)] = 3 [sin (52.5° + 127.5°) - sin (52.5° - 127.5°)]
=2 [sin (180°) - sin (-75°)]
= 3 [sin 180°+ sin 75°] {Since, sin (-θ) = - sin θ}
= 3 [1 + 0.9659] {Since, sin 35° = 0.5735, sin 180° = 0}
= 5.8977
Hence, 6 sin (52.5 °) sin (127.5°) = 5.8977
Example 7: Use the identity to simplify 2 cos (7x) sin (3x).
Solution: From 2cosasinb formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
Now, 2 cos 7x sin 3x = sin (7x + 3x) - sin (7x - 3x)
= sin 10x - sin 4x
Hence, 2 cos 7x sin 3x = sin 10x - sin 4x
Example 8: Given that A=90o and B=45, find the value of 2cosAsinB and verify it using formula.
Solution: From 2cosAsinB formula we have,
2 cos A sin B = sin (A + B) - sin (A - B)
Now substituting the values: 2 cos 90 sin 45 = sin(90+45) - sin(90-45)
2 cos 90 sin 45 = sin 135 - sin 45
Since cos 90 = 0, the Left hand side(LHS) becomes 0.
Now right hand side(RHS): sin 135 = 1/√2, sin 45 = 1/√2
so, sin 135 - sin 45 = 0
The identity is thus verified.
Example 9 : Prove that sin 4x - sin 2x = 2 cos 3x sin x
Solution: Start with the right hand side and use identity:
2 cos 3x sin x = sin(3x+x)-sin(3x-x)
2 cos 3x sin x = sin 4x - sin 2x
Hence proved.
Example 10: Simplify the expression 2 cos(5x) sin(7x)
Solution: Using the identity:
2cos(5x)sin(7x) = sin(5x+7x) - sin(5x-7x)
2cos(5x)sin(7x) = sin 12x + sin 2x
Conclusion
The formula 2cosAsinB is a key identity in trigonometry that simplifies many complex expressions. With regular practice, as provided in the above examples, will reinforce your understanding an ability to apply this identity effectively.
All in all, this article aims to provide a comprehensive understanding of the formula 2cosAsinB , its derivation, and applications.