2 Cos A Sin B

Sin a Cos b

Sin a cos b is an important trigonometric identity that is used to solve complicated problems in trigonometry. Sin a cos b is used to obtain the production of the sine function of bending a and cosine function of bending b. Information technology can be obtained from angle sum and angle deviation identities of the sine role. sin a cos b formula is written as (one/2)[sin(a+b) + sin(a-b)].

In this article, we volition explore the sin a cos b formula, its proof, and acquire its application to solve diverse trigonometric problems with the help of solved examples.

1.	What is Sin a Cos b Identity?
2.	Proof of Sin a Cos b Formula
3.	Application of Sin a Cos b Identity
four.	FAQs on Sin a Cos b

What is Sin a Cos b Identity?

Sin a cos b is a trigonometric identity used to solve various bug in trigonometry. Sin a cos b is equal to half the sum of sine of the sum of angles a and b, and sine of divergence of angles a and b. Mathematically, it is written as sin a cos b = (1/2)[sin(a + b) + sin(a - b)], that is, it tin can be derived using the trigonometric identities sin (a + b) and sin(a - b). sin a cos b formula can be applied when the sum and difference of angles a and b are known, or when 2 angles a and b are known.

Sin a Cos b Formula

The formula for sin a cos b is given by, sin a cos b = (1/2)[sin(a + b) + sin(a - b)]. The formula for sin a cos b can be applied when the chemical compound angles (a + b) and (a - b) are known, or when values of angles a and b are known.

sin a cos b formula

Proof of Sin a Cos b Formula

Now that we know the formula of sin a cos b, which is sin a cos b = (one/2)[sin(a + b) + sin(a - b)], nosotros will derive this formula using the trigonometric formulas and identities. Sin a cos b formula tin can be derived using the angle sum and angle difference formulas of the sine part. We volition utilize the following trigonometric formulas:

sin (a + b) = sin a cos b + cos a sin b --- (ane)
sin (a - b) = sin a cos b - cos a sin b --- (2)

Adding equations (1) and (two), we have

sin (a + b) + sin (a - b) = (sin a cos b + cos a sin b) + (sin a cos b - cos a sin b) (From (1) and (2))

⇒ 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 + sin a cos b) + (cos a sin b - cos a sin b)

⇒ sin (a + b) + sin (a - b) = 2 sin a cos b + 0

⇒ sin (a + b) + sin (a - b) = 2 sin a cos b

⇒ sin a cos b = (1/2) [sin (a + b) + sin (a - b)]

Hence, we accept obtained the sin a cos b formula using the sin (a + b) and sin (a - b) identities.

Awarding of Sin a Cos b Identity

Since we have derived the sin a cos b formula, now we will larn how to apply the formula to solve simple trigonometric and integration problems. We will consider some examples based on sin a cos b identity and solve them pace-wise. Permit us understand the application of the sin a cos b formula by following the given steps:

Example 1: Express the trigonometric office sin 7x cos 3x as a sum of the sine part.

Step i: Nosotros will use the sin a cos b formula: sin a cos b = (1/2) [sin (a + b) + sin (a - b)]. Identify the values of a and b in the formula. We have sin 7x cos 3x, here a = 7x, b = 3x.

Step ii: Substitute the values of a and b in the formula sin a cos b = (one/2) [sin (a + b) + sin (a - b)]

sin 7x cos 3x = (1/2) [sin (7x + 3x) + sin (7x - 3x)]

⇒ sin 7x cos 3x = (1/2) [sin (10x) + sin (4x)]

⇒ sin 7x cos 3x = (1/2) sin (10x) + (1/2) sin (4x)

Hence, nosotros can write sin 7x cos 3x every bit (ane/2) sin (10x) + (one/2) sin (4x) as a sum of sine function.

Instance ii: Evaluate the integral ∫sin 2x cos 4x dx using the sin a cos b formula.

Footstep ane: First, we will express sin 2x cos 4x as a sum of sine function using the formula sin a cos b = sin a cos b = (ane/2) [sin (a + b) + sin (a - b)]. Identify a and b in sin 2x cos 4x. We accept a = 2x, b = 4x.

Stride 2: Substitute the values of a and b in the formula sin a cos b = (1/2) [sin (a + b) + sin (a - b)]

sin 2x cos 4x = (1/two) [sin (2x + 4x) + sin (2x - 4x)]

⇒ sin 2x cos 4x = (1/two) [sin (6x) + sin (-2x)]

⇒ sin 2x cos 4x = (i/2) sin (6x) - (ane/two) sin (2x) [Because sin(-a) = -sin a]

Step 3: Substitute sin 2x cos 4x = (1/ii) sin (6x) - (i/2) sin (2x) into the integral ∫sin 2x cos 4x dx.

∫sin 2x cos 4x dx = ∫ [(1/ii) sin (6x) - (one/two) sin (2x)] dx

⇒ ∫sin 2x cos 4x dx = (1/2) ∫sin(6x) dx - (1/2) ∫sin(2x) dx

⇒ ∫sin 2x cos 4x dx = (1/2)[-cos(6x)]/vi - (1/2)[-cos(2x)]/2 + C

⇒ ∫sin 2x cos 4x dx = (-1/12) cos (6x) + (1/4) cos (2x) + C

Hence, nosotros take solved the integral ∫sin 2x cos 4x dx using sin a cos b formula and is equal to (-1/12) cos (6x) + (1/4) cos (2x) + C.

Important Notes on Sin a Cos b

sin a cos b = (1/2)[sin(a+b) + sin(a-b)]
sin a cos b formula is practical when angles a and b are known, or when the sum and difference of angles a and b are known.
sin a cos b formula is used to solve simple and complex trigonometric problems.
Sin a cos b is equal to half the sum of sine of the sum of angles a and b, and sine of deviation of angles a and b.

Related Topics on Sin a Cos b

sin a sin b
cos a cos b
sin of ii pi
cos 2x

Sin a Cos b Examples

Example ane: Evaluate the integral ∫sin 12x cos 3x dx using the sin a cos b formula.

Solution: First, nosotros will express sin 12x cos 3x every bit a sum of sine office using the formula sin a cos b = sin a cos b = (1/2) [sin (a + b) + sin (a - b)]. Identify a and b in sin 12x cos 3x. We have a = 12x, b = 3x.

Substitute the values of a and b in the formula sin a cos b = (1/2) [sin (a + b) + sin (a - b)]

sin 12x cos 3x = (1/two) [sin (12x + 3x) + sin (12x - 3x)]

⇒ sin 12x cos 3x = (1/2) [sin (15x) + sin (9x)]

⇒ sin 12x cos 3x = (1/2) sin (15x) + (i/ii) sin (9x)

Substitute sin 12x cos 3x = (1/2) sin (15x) + (ane/2) sin (9x) into the integral ∫sin 12x cos 3x dx.

∫sin 12x cos 3x dx = ∫ [(1/2) sin (15x) + (i/2) sin (9x)] dx

⇒ ∫sin 12x cos 3x dx = (1/2) ∫sin(15x) dx + (1/two) ∫sin(9x) dx

⇒ ∫sin 12x cos 3x dx = (1/two)[-cos(15x)]/15 + (1/2)[-cos(9x)]/9 + C

⇒ ∫sin 12x cos 3x dx = (-1/30) cos (15x) + (-1/18) cos (9x) + C

Hence, nosotros accept solved the integral ∫sin 12x cos 3x dx using sin a cos b formula and is equal to (-1/thirty) cos (15x) + (-1/18) cos (9x) + C.

Respond: ∫sin 12x cos 3x dx = (-1/30) cos (15x) + (-1/eighteen) cos (9x) + C
Example ii: Limited the trigonometric function sin 3x cos 9x as a sum of the sine office using sin a cos b formula.

Solution: We will use the sin a cos b formula: sin a cos b = (1/ii) [sin (a + b) + sin (a - b)]. Identify the values of a and b in the formula. Nosotros accept sin 3x cos 9x, hither a = 3x, b = 9x.

Substitute the values of a and b in the formula sin a cos b = (one/ii) [sin (a + b) + sin (a - b)]

sin 3x cos 9x = (1/2) [sin (3x + 9x) + sin (3x - 9x)]

⇒ sin 3x cos 9x = (i/two) [sin (12x) + sin (-6x)]

⇒ sin 3x cos 9x = (one/two) sin (12x) - (1/2) sin (6x) [Because sin(-a) = -sin a]

Hence, we can write sin 3x cos 9x as (1/2) sin (12x) - (i/2) sin (6x) equally a sum of sine function using the sin a cos b formula.

Answer: sin 3x cos 9x = (one/2) sin (12x) - (ane/2) sin (6x)

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Do Questions on Sin a Cos b

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FAQs on Sin a Cos b

What is Sin a Cos b in Trigonometry?

Sin a cos b is an important trigonometric identity that is used to solve complicated bug in trigonometry given by sin a cos b = (i/ii) [sin (a + b) + sin (a - b)]

What is the Formula of Sin a Cos b?

The formula of sin a cos b is sin a cos b = (1/ii) [sin (a + b) + sin (a - b)]

What is the Formula of 2 sin a cos b?

The formula for 2 sin a cos b is given by, 2 sin a cos b = sin (a + b) + sin (a - b)

Discover the Exact Value of sin a cos b when a = 90° and b = 180°.

Substitute a = xc° and b = 180° in sin a cos b = (1/2) [sin (a + b) + sin (a - b)]. sin 90° cos 180° = (1/2) [sin (90° + 180°) + sin (ninety° - 180°)] = (one/2) [sin 270° + sin(-90°)] = (ane/2)(-1-1) = -1. Hence, sin a cos b = -1 when a = 90° and b = 180°

How to Find sin a cos b formula?

Sin a Cos b formula can be calculated using sin(a + b) and sin (a - b) trigonometric identities.

When is sin a cos b equal to (i/2) sin 2a?

sin a cos b is equal to (ane/2) sin 2a when a = b. When a = b in sin a cos b = (1/2) [sin (a + b) + sin (a - b)], nosotros have sin a cos b = (1/2) [sin (a + a) + sin (a - a)] = (1/2) [sin 2a + 0] = (i/ii) sin 2a.

How to Evidence sin a cos b Identity?

Sin a cos b formula can be proved using the angle sum and angle difference formulas of the sine function.

What is the Expansion of Sin a Cos b?

The expansion of sin a cos b is given by sin a cos b = (1/2) [sin (a + b) + sin (a - b)].

What is the Difference Between Sin a Cos b Formula and Cos a Sin b Formula?

Sin a cos b formula is the sum of sin (a + b) and sin (a - b) trigonometric identities, whereas cos a sin b formula is the difference of sin (a + b) and sin (a - b) trigonometric identities, that is, sin a cos b = (1/2) [sin (a + b) + sin (a - b)] and cos a sin b = (ane/2) [sin (a + b) - sin (a - b)].

2 Cos A Sin B,

Source: https://www.cuemath.com/trigonometry/sina-cosb/

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