How to find the value of sin4x, cos4x, cot4x?

The value of sin4x, cos4x and cot4x can be found out by using the basic trigonometric formulas.

We will use the basic trigonometric formulas to find the values of sin4x, cos4x, and cot4x.

Solution:

sin 4x = sin2(2x)

We know that, sin 2x = 2 sinx cosx

Thus, sin2(2x) = 2 sin2x cos2x

= 2 (2 sinx cosx) cos2x

= 4 sinx cosx (2cos²x - 1) [ Since, cos2x = 2cos²x - 1]

= 8 sinx cos³x - 4 sinx cosx

Thus, sin 4x = 8 sinx cos³x - 4 sinx cosx

Now, solving for cos4x we have,

cos 4x = cos2(2x)

We know that,

cos2x = 2cos²x - 1

Thus, cos2(2x) = 2cos²(2x) - 1

= 2 (2cos²x - 1) -1

= 4 cos²x - 2 + 1

= 4 cos²x - 1

Thus, cos 4x = 4 cos²x - 1

Now, solving for cot 4x,

cot 4x = cot2(2x)

We know that,

cot 2x = (cot²x - 1) / 2 cotx

Thus, cot 2(2x) = [cot²(2x) - 1] / 2 cot(2x)

= [(cot²x - 1) / 2 cotx]² / [ 2 {(cot²x - 1) / 2 cotx}]

On solving,

= (cot²x - 1) / 4

Thus, cot 4x = (cot²x - 1) / 4

Hence, the value of sin4x is 8 sinx cos³x - 4 sinx cosx, the value of cos4x is 4 cos²x - 1, and the value of cot4x is (cot²x - 1) / 4.

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How to find the value of sin4x, cos4x, cot4x?

Summary:

The value of sin4x is 8 sinx cos³x - 4 sinx cosx, the value of cos4x is 4 cos²x - 1, and the value of cot4x is (cot²x - 1) / 4.