Sin 2x Cos 2x

Sin 2x cos 2x is one of the trigonometric identities which is essential for solving a variety of trigonometry related questions. Here, the simplified value of Sin2x cos2x is given along with the integral and derivative of sin2x and cos 2x.

What is the Value of Sin 2x Cos 2x?

The value of sin 2x × Cos 2x is:

 Sin 2x Cos 2x = 2 Cos x (2 Sin x Cos2 x − Sin x)Or, Sin 2x Cos 2x = 2 Cos x (Sin x – 2 Sin3 x)

How to Derive Sin 2x Cos 2x Value?

To find the value of sin2x × Cos 2x, the trigonometric double angle formulas are used. For the derivation, the values of sin 2x and cos 2x are used.

From trigonometric double angle formulas,

Sin 2x = 2 sin x cos x ————(i)

And,

Cos 2x = Cos2x − Sin2x

= 2 cos2x − 1 ————(ii) [Since Sin2 x + Cos2 x = 1]

= 1 − 2Sin2x ————(iii)

Also Check: Trigonometry

Now, to get the value of Sin 2x Cos 2x, multiply equation (i) with (ii) or (i)

Consider equation (i) and (ii),

Sin 2x = 2 sin x cos x

And,

Cos 2x = 2 cos2x − 1

Multiply them to get,

Sin 2x Cos 2x = 2 Sin x Cos x (2 cos2x − 1)

= 4 Sin x Cos3 x − 2 Sin x Cos x

= 2 Cos x (2 Sin x Cos2 x − Sin x)

Now, consider equation (i) and (iii),

Sin 2x = 2 sin x cos x

And,

Cos 2x = 1 − 2 Sin2x

Multiply them to get,

Sin 2x Cos 2x = 2 Sin x Cos x (1 − 2 Sin2x)

= 2 Sin x Cos x − 4 Sin3 x Cos x

= 2 Cos x (Sin x – 2 Sin3 x)

So,

• Sin 2x Cos 2x = 2 Cos x (2 Sin x Cos2 x − Sin x)

Or,

• Sin 2x Cos 2x = 2 Cos x (Sin x – 2 Sin3 x)

Derivative of Sin 2x Cos 2x

 d/dx (Sin 2x Cos 2x) = 2Cos(4x)

Proof:

Sin(2x)cos(2x)

= ½(2sin(2x)cos(2x))

Or, ½Sin(4x)

Now, differentiate the given function w.r.t. x:

d/dx [½Sin(4x)]

= ½[d/dx(Sin(4x))]

= ½[Cos(4x)d/dx(4x)]

= ½[Cos(4x)(4)]

So, d/dx (Sin 2x Cos 2x) = 2 Cos(4x)

Integral of Sin 2x Cos 2x

 ∫ (Sin 2x Cos 2x) = (Sin 2x)2/ 4 + C

Proof:

Consider sin 2x = u

So, du/dx = 2Cos(2x)

Or, dx = du/2Cos(2x)

Now, ∫u Cos(2x)dx = ∫u • Cos(2x) • du/2cos 2x

Here, Cos 2x can be cancelled out.

So,

∫u Cos(2x)dx = ∫(u • du/2)

= ½[∫u du]

= ½ u2/2 + c

= u2/4 + C

Or, ∫ (Sin 2x Cos 2x) = (Sin 2x)2/ 4 + C

Sin 2x Cos 2x Examples

Example 1:

If sin θ + sin 2θ + sin 3θ = sin α, cos θ + cos 2θ + cos 3θ = cos α, then theta is equal to

Solution:

Given,

sin θ + sin 2θ + sin 3θ = sin α

The given equation can be written as:

sin 2θ(2 cos θ + 1) = sin α….(1)

Also, given:

cos θ + cos 2θ + cos 3θ = cos α

Similarly, the above equation can be written as:

cos 2θ(2 cos θ + 1) = cos α….(2)

Now, divide (1) by (2), we get

[sin 2θ(2 cos θ + 1)]/ [cos 2θ(2 cos θ + 1)] = sin α/cos α

tan 2θ = tan α

2θ = α

θ = α/2

Example 2:

Find the value of (sin 8x + 7sin 6x + 18 sin 4x + 12 sin 2x)/ (sin 7x+6 sin 5x+12 sin 3x).

Solution:

Given expression: (sin 8x + 7sin 6x + 18 sin 4x + 12 sin 2x)/ (sin 7x+6 sin 5x+12 sin 3x).

The numerator of the given expression is sin 8x + 7sin 6x + 18 sin 4x + 12 sin 2x.

Now, simplify the expression:

= (sin 8x + sin6x) + 6(sin 6x + sin 4x) + 12(sin 4x + sin 2x)

= 2sin7x cosx + 12sin5x cosx + 24sin3x cos x

= 2cos x (sin7x + 6sin5x + 12sin3x)

Substituting the simplified numerator expression in the given expression, we get

(sin 8x + 7sin 6x + 18 sin 4x + 12 sin 2x)/ (sin 7x+6 sin 5x+12 sin 3x)=2cos x (sin7x + 6sin5x + 12sin3x)/(sin 7x+6 sin 5x+12 sin 3x)

= 2 cos x.

Hence, the value of (sin 8x + 7sin 6x + 18 sin 4x + 12 sin 2x)/ (sin 7x+6 sin 5x+12 sin 3x) is  2 cos x.

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