Power-Reducing Formulas and How to Use Them (With Examples)

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The power-reducing formula is an identity useful in rewriting trigonometric functions raised to powers. These identities are rearranged double-angle identities that function much like the double-angle and half-angle formulas.

Power-reducing identities in Calculus are useful in simplifying equations that contain trigonometric powers resulting in reduced expressions without the exponent. Reducing the power of the trigonometric equations gives more space to understand the relationship between the function and its rate of change every single time. It may be any trig function such as sine, cosine, tangent, or their inverses raised to any power.

For instance, the given problem is a trigonometric function raised to the fourth power or higher; it can apply the power-reducing formula more than once to eliminate all the exponents until fully reduced.

Power-Reducing Formulas for Squares

sin² (u) = (1 – cos (2u)) / 2

cos² (u) = (1 + cos (2u)) / 2

tan² (u) = (1 – cos (2u)) / (1 + cos (2u))

Power-Reducing Formulas for Cubes

sin³ (u) = (3sin (u) – sin (3u)) / 4

cos³ (u) = (3cos (u) – cos (3u)) / 4

tan³ (u) = (3sin (u) – sin (3u)) / (3cos (u) – cos (3u))

Power-Reducing Formulas for Fourths

sin⁴ (u) = [3 – 4 cos (2u) + cos (4u)] / 8

cos⁴ (u) = [3 + 4 cos (2u) + cos (4u)] / 8

tan⁴ (u) = [3 – 4 cos (2u) + cos (4u)] / [3 + 4 cos (2u) + cos (4u)]

Power-Reducing Formulas for Fifths

sin⁵ (u) = [10 sin (u) – 5 sin (3u) + sin (5u)] / 16

cos⁵ (u) = [10 cos (u) + 5 cos (3u) + cos (5u)] / 16

tan⁵ (u) = [10 sin (u) – 5 sin (3u) + sin (5u)] / [10 cos (u) + 5 cos (3u) + cos (5u)]

Special Power-Reducing Formulas

sin² (u) cos² (u) = (1 – cos (4u)) / 8

sin³ (u) cos³ (u) = (3 sin (2u) – sin (6u)) / 32

sin⁴ (u) cos⁴ (u) = (3 - 4 cos (4u) + cos (8u)) / 128

sin⁵ (u) cos⁵ (u) = (10 sin (2u) - 5 sin (6u) + sin (10u)) / 512

Power-Reducing Formula Proof

The power reduction formulas are further derivations of the double angle, half-angle, and the Pythagorean Identify. Recall the Pythagorean equation shown below.

sin² (u) + cos² (u) = 1

Let us first prove the power reducing formula for sine. Recall that the double angle formula cos (2u) is equal to 2 cos² (u) – 1.

(1 – cos 2u) / 2 = [1 – (2 cos² (u) – 1)] / 2

(1 – cos 2u) / 2 = [2(1 - cos² (u)] / 2

(1 – cos 2u) / 2 = 1 – cos² (u)

1 – cos² (u) = sin² (u)

Next, let us prove the power reducing formula for cosine. Still considering that the double angle formula cos (2u) is equal to 2 cos² (u) – 1.

(1 + cos 2u) / 2 = [1 + (2 cos² (u) – 1)] / 2

(1 + cos 2u) / 2 = [2(cos² (u))] / 2

(1 + cos 2u) / 2 = cos² (u)

Example 1: Using Power-Reducing Formulas for Sine Functions

Find the value of sin⁴x given that cos(2x) = 1/5.

Solution

Since the given sine function has an exponent to the fourth power, express the equation sin⁴x as a squared term. It will be much easier to write the fourth power of the sine function in terms of squared power to avoid the use of the half-angle identities and double-angle identities.

sin⁴ (x) = (sin²x)²

sin⁴ (x) = ((1 – cos (2x))/2)²

Substitute the value of cos (2x) = 1/5 to the squared power reduction rule for the sine function. Then, simplify the equation to get the result.

sin⁴ (x) = ((1 – 1/5)/2)²

sin⁴ (x) = 4/25

Final Answer

The value of sin⁴x given that cos (2x) = 1/5 is 4/25.

Example 2: Rewriting a Sine Equation to the Fourth Power Using the Power-Reducing Identities

Rewrite the sine function sin⁴x as an expression without powers larger than one. Express it in terms of the first power of the cosine.

Solution

Simplify the solution by writing the fourth power in terms of squared power. Although it can be expressed as (sin x)(sin x)(sin x)(sin x), but remember to retain at least a squared power in order to apply the identity.

sin⁴x = (sin²x)²

Use the power-reducing formula for cosine.

sin⁴x = ((1 – cos (2x))/2)²

sin⁴x = (1 – 2 cos (2x) + cos² (2x)) / 4

Simplify the equation to its reduced form.

sin⁴x = (1/4) [1 – 2cos (2x) + (1 + cos (4x)/2)]

sin⁴x = (1/4) – (1/2) cos 2x + 1/8 + (1/8) cos 4x

sin⁴x = (3/8) – (1/2) cos 2x + (1/8) cos 4x

Final Answer

The reduced form of the equation sin⁴x is (3/8) – (1/2) cos 2x + (1/8) cos 4x.

Example 3: Simplifying Trigonometric Functions to the Fourth Power

Simplify the expression sin⁴ (x) – cos⁴ (x) using the power-reducing identities.

Solution

Simplify the expression by reducing the expression into square powers.

sin⁴ (x) – cos⁴ (x) = (sin² (x) – cos² (x)) (sin² (x) + cos² (x))

sin⁴ (x) – cos⁴ (x) = - (cos² (x) – sin² (x))

Apply the double angle identity for cosine.

sin⁴ (x) – cos⁴ (x) = - cos (2x)

Final Answer

The simplified expression of sin⁴ (x) – cos⁴ (x) is - cos (2x).

Example 4: Simplifying Equations to Sines and Cosines of First Power

Using the power-reduction identities, express the equation cos² (θ) sin² (θ) using only cosines and sines to the first power.

Solution

Apply the power-reducing formulas for cosine and sine, and multiply both of them. See the following solution below.

cos²θ sin²θ = cos² (θ) sin² (θ)

cos²θ sin²θ = (1/4) (2 cos θ sin θ)²

cos²θ sin²θ = (1/4) (sin² (2θ))

cos²θ sin²θ = (1/4) [(1 – cos (4θ))/2]

cos²θ sin²θ = (1/8) [1 – cos (4θ)]

Final Answer

Therefore, cos² (θ) sin² (θ) = (1/8) [1 – cos (4θ)].

Example 5: Proving the Power Reducing-Formula for Sine

Prove the power-reducing identity for sine.

sin²x = (1 – cos (2x)) / 2

Solution

Start simplifying the double-angle identity for cosine. Remember that cos (2x) = cos² (x) – sin² (x).

cos (2x) = cos² (x) – sin² (x)

cos (2x) = (1 – sin² (x)) – sin² (x)

cos (2x) = 1 – 2 sin² (x)

Use the double-angle identity to simplify sin² (2x). Transpose 2 sin² (x) to the left equation.

2 sin² (x) = 1 – cos (2x)

sin² (x) = [(1 – cos (2x)) / 2]

Final Answer

Therefore, sin² (x) = [(1 – cos (2x)) / 2].

Example 6: Solving the Value of a Sine Function Using Power-Reducing Formula

Solve the sine function sin² (25°) using the power-reducing identity for sine.

Solution

Recall the power-reducing formula for sine. Then, substitute the value of the angle measure u = 25° to the equation.

sin² (x) = [(1 – cos (2x)) / 2]

sin² (25°) = [(1 – cos (2(25°))) / 2]

Simplify the equation and solve for the resulting value.

sin² (25°) = [(1 – cos (50°)) / 2]

sin² (25°) = 0.1786

Final Answer

The value of sin² (25°) is 0.1786.

Example 7: Expressing the Fourth Power of Cosine to the First Power

Express the power-reducing identity cos⁴ (θ) using only sines and cosines to the first power.

Solution

Apply the formula for cos² (θ) two times. Consider θ as x.

cos⁴ (θ) = (cos² (θ))²

cos⁴ (θ) = ([(1 + cos (2θ)]/2)²

Square both the numerator and the denominator. Use the power-reducing formula for cos² (θ) with θ = 2x.

cos⁴ (θ) = [(1 + 2 cos (2θ) + cos² (2θ)] / 4

cos⁴ (θ) = [(1 + 2 cos (2θ) + [1 + cos (4θ)/2]] / 4

cos⁴ (θ) = [(2 + 4 cos (2θ) + 1 + cos (4θ)] / 8

Simplify the equation and distribute 1/8 through the parentheses

cos⁴ (θ) = (1/8) [cos (4θ) + 4 cos (2θ) + 3)

cos⁴ (θ) = (1/8) cos (4θ) + (1/2) cos (2θ) + 3/8

Final Answer

The expression for cos⁴ (θ) with only sines and cosines to the first power is (1/8) cos (4θ) + (1/2) cos (2θ) + 3/8.

Example 8: Proving Equations Using Power-Reducing Formula

Use the power-reducing formula to prove that 5 cos⁴ (x) = 15/8 + (5/2) cos (2x) + (5/8) cos (4x).

Solution

Rewrite the equation and apply the formula for cos² (x) two times. Consider θ as x.

5 cos⁴ (x) = 5 (cos² (x))²

Substitute the reduction formula for cos² (x). Raise both the denominator and numerator the dual power.

5 cos⁴ (x) = 5 [(1 + cos (2x)) / 2]²

5 cos⁴ (x) = (5/4) [1 + 2 cos (2x) + cos² (2x)]

Substitute the power-reducing formula of cosine to the last term of the resulting equation.

5 cos⁴ (x) = (5/4) + (5/2) cos (2x) + (5/4) [(1 + cos 2 (2x)) / 2]

5 cos⁴ (x) = (5/4) + (5/2) cos (2x) + (5/8) + (5/8) cos (4x)

5 cos⁴ (x) = 15/8 + (5/2) cos (2x) + (5/8) cos (4x)

Final Answer

Therefore, 5 cos⁴ (x) = 15/8 + (5/2) cos (2x) + (5/8) cos (4x).

Example 9: Proving Identities Using the Power-Reducing Formula for Sine

Prove that sin³ (3x) = (1/2) [sin (3x)] [1 – cos (3x)].

Solution

Since the trigonometric function is raised to the third power, there will be one quantity of square power. Rearrange the expression and multiply one square power to a single power.

sin³ (3x) = [sin (3x)] [sin² (3x)]

Substitute the power-reduction formula to the obtained equation.

sin³ (3x) = [sin (3x)] [(1 – cos (3x))/2]

Simplify to its reduced form.

sin³ (3x) = sin (3x) (1/2) (1 – cos (3x))

sin³ (3x) = (1/2) [sin (3x)] [1 – cos (3x)]

Final Answer

Therefore, sin³ (3x) = (1/2) [sin (3x)] [1 – cos (3x)].

Example 10: Rewriting a Trigonometric Expression Using the Power-Reducing Formula

Rewrite the trigonometric equation 6sin⁴ (x) as an equivalent equation having no powers of functions larger than 1.

Solution

Start rewriting sin² (x) to another power. Apply the power-reduction formula two times.

6 sin⁴ (x) = 6 [sin² (x)]²

Substitute the power-reducing formula for sin² (x).

6 sin⁴ (x) = 6 [(1 – cos (2x)) / 2]²

Simplify the equation by multiplying and distributing constant 3/2.

6 sin⁴ (x) = 6 [(1 – 2 cos (2x) + cos² (2x))] / 4

6 sin⁴ (x) = (3/2) [1 – 2 cos (2x) + cos² (2x)]

6 sin⁴ (x) = (3/2) – 3 cos (2x) + (3/2) cos² (2x)

Final Answer

Therefore, 6 sin⁴ (x) is equal to (3/2) – 3 cos (2x) + (3/2) cos² (2x).

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