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 2: Rewriting a Sine Equation to the Fourth Power Using the Power-Reducing Identities
John Ray Cuevas
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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