# 2sinAsinB Formula

• Last Updated : 27 Jun, 2022

2sinasinb is one of the important trigonometric formulas which is equal to cos (a – b) – cos (a + b). In mathematics, trigonometry is an important branch that deals with the relationship between angles and sides of a right-angled triangle, which has its applications in various fields like astronomy, aviation, marine biology, astronomy, etc. There are six trigonometric ratios, of which three ratios are the reciprocals of the other three trigonometric ratios. A trigonometric ratio is a ratio between the lengths of the sides of a right triangle. Trigonometric Ratios

1. sin θ = opposite side/hypotenuse
2. cos θ = adjacent side/hypotenuse
3. tan θ = opposite side/adjacent side
4. cosec θ = 1/sin θ = hypotenuse/opposite side
5. sec θ = 1/cos θ = hypotenuse/adjacent side
6. cot θ = 1/tan θ = adjacent side/opposite side

### 2sinasinb formula

The 2sinasinb formula is a trigonometric formula that is used to simplify trigonometric expressions and also solve complex integrals and derivatives of trigonometric expressions. The 2sinasinb formula is equal to the difference between the angle sum and the angle difference of the cosine functions, i.e., for two angles A and B,
2 sin A sin B = cos (A-B) – cos (A + B).

The 2sinasinb formula is,

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

From the formula, we can observe that twice the product of two sine functions is converted into the difference between the angle sum and the angle difference of the cosine functions. With the help of the 2 sin A sin B formula, we can extract the formula of sin A sin B.

sin A sin B = ½ [cos (A – B) – cos (A + B)]

### Derivation of 2sinasinb formula

We can derive the 2sinasinb formula with the help of the sum and difference of formulae of the cosine function.

cos (A + B) = cos A cos B – sin A sin B    ———— (1)

cos (A – B) = cos A cos B + sin A sin B    ———— (2)

Now subtract the equation(1) from the equation (2)

⇒ cos (A – B) – cos (A + B) = [cos A cos B + sin A sin B] – [cos A cos B – sin A sin B]

⇒ cos (A – B) – cos (A + B) = cos A cos B + sin A sin B – cos A cos B + sin A sin B

⇒ cos (A – B) – cos (A + B) = sin A sin B + sin A sin B

⇒ cos (A – B) – cos (A + B) = 2 sin A sin B

Hence, 2 sin A sin B = cos (A – B) – cos (A + B)

cos 2A formula

Let us assume that A = B,

We have, 2 sin A sin B = cos (A – B) – cos (A + B)

By substituting A = B in the above equation, we get

⇒ 2 sin A sin A = cos (A – A) – cos (A + A)

⇒ 2 sin2 A = cos (0°) – cos 2A

⇒ 2 sin2 A = 1 – cos 2A           {Since, cos 0° = 1}

⇒ cos 2A = 1 – 2sin2 A

cos 2A = 1 – 2sin2 A

### Sample Problems

Problem 1: Solve the integral of 3 sin 5x sin (9x/2).

Solution:

Integral of 3 sin 5x sin (9x/2) = ∫3 sin 5x sin (9x/2) dx

From 2sinasinb formula we have,

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

3 sin 5x sin (9x/2) = 3/2 [2 sin 5x sin (9x/2)]

= 3/2 [cos (5x – (9x/2)) – cos (5x + (9x/2))]

= 3/2 [cos (x/2) – cos (19x/2)]

Now, ∫3 sin 5x sin (9x/2) dx = ∫3/2 [cos(x/2) – cos(19x/2)] dx

= 3/2 ∫cos(x/2) dx – ∫cos(19x/2) dx

= 3/2 [2 sin (x/2) – 2/19 sin (19x/2)]    {∫cos (ax) = 1/a sin (ax) + c}

= 3 sin (x/2) – 3/19 sin (19x/2)

Hence, the  integral of 3 sin 5x sin (9x/2) = 3 [sin(x/2) – 1/19 sin (19x/2)]

Problem 2: Find the derivative of 4 sin 2x sin (5x/2).

Solution:

Derivative of 4 sin 2x sin (5x/2) = d(4 sin 2x sin (5x/2))/dx

From 2sinasinb formula we have,

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

Now, 4 sin 2x sin (5x/2) = 2 [2 sin 2x sin (5x/2)]

= 2 [cos (2x – (5x/2)) – cos (2x + (5x/2))]

= 2 [cos (-x/2) – cos (9x/2)]

= 2[cos (x/2) – cos (9x/2)]      {Since, cos (-θ) = cos θ}

Now, d(4 sin 2x sin (5x/2))/dx = d{2[cos (x/2) – cos (9x/2)]}/dx

= 2{d(cos(x/2))/dx – d(cos(9x/2))/dx}

= 2{1/2 (-sin (x/2) – (- (9/2) sin (9x/2)}      {Since, d(cos ax)/dx = – a sin ax}

= 9 sin (9x/2) – sin (x/2)

Hence, the derivative of 4 sin 2x sin (5x/2) = 9 sin (9x/2) – sin (x/2).

Problem 3: Express 6 sin (11x/2) sin 7x in terms of the cosine function.

Solution:

From 2sinasinb formula we have,

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

Now, 6 sin (11x/2) sin 7x = 3 [2 sin (11x/2) sin 7x]

= 3 [cos (11x/2 – 7x) – cos (11x/2 + 7)]

= 3 [cos (-3x/2) – cos (25x/2)]

= 3 [cos (3x/2) – cos (25x/2)]        {Since, cos (-θ) = cos θ}

Hence, 6 sin (11x/2) sin 7x = 3 [cos (3x/2) – cos (25x/2)].

Problem 4: Find the value of the expression 7 sin (17.5°) sin (72.5°) using the 2sinasinb formula.

Solution:

7 sin (17.5°) sin (72.5°) = 7/2 [2 sin (17.5°) sin (72.5°)]

From 2sinasinb formula we have,

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

Now, 7/2 [2 sin (17.5°) sin (72.5°)] = 7/2 [cos (17.7° – 72.5°) – cos (17.5° + 72.5°)]

= 7/2 [cos (-55°) – cos (90°)]

= 7/2 [cos 55° – cos 90°]      {Since, cos (-θ) = cos θ}

= 7/2 [0.5735 – 0]               {Since, cos 55° = 0.5735, cos 90° = 0}

= 2.00725‬

Hence, 7 sin (17.5°) sin (72.5°) = 2.00725‬

Problem 5: Express 2 sin (12x) sin (17x/2) in terms of the cosine function.

Solution:

From 2sinasinb formula we have,

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

Now, 2 sin (12x) sin (17x/2) = [cos (12x – (17x/2) – cos(12x + (17x/2)]

= cos [(24x – 17x)/2] – cos [(24x + 17x)/2]

= cos (7x/2) – cos (41x/2)

Hence, 2 sin (12x) sin (17x/2) = cos (7x/2) – cos (41x/2).

Problem 6: Solve 4 sin (66.5°) sin (113.5°) using the 2sinasinb formula.

Solution:

From 2sinasinb formula we have,

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

Now, 4 sin (66.5°) sin (113.5°) = 2 [2 sin (66.5°) sin (113.5°)]

= 2 [cos (66.5° – 113.5°) – cos (66.5° + 113.5°)]

= 2 [cos (-47°) – cos (180°)]

= 2 [cos 47° – cos 180°]        {Since, cos (-θ) = cos θ}

= 2 [0.682 – 1]                       {Since, cos 47° = 0.682, cos 180° = -1}

= -0.636

Hence, 4 sin (66.5°) sin (113.5°) = -0.636

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