use power reducing identities to prove the identity
sin^4x=1/8(3-4cos2x+cos4x)
cos^3x=(1/2cosx) (1+cos2x)
thanks :)
cos 2x = 2cos^2 x - 1
so, 1/2 cos x (1+2cos^2 x - 1) = cos^3 x
cos 4x = 1 - 2sin^2 2x
= 1 - 8sin^2 x cos^2 x
= 1 - 8sin^2 x (1 - sin^2 x)
= 1 - 8sin^2 x + 8 sin^4 x
cos 2x = 1 - 2sin^2 x
4cos 2x = 4 - 8sin^2 x
1/8(3-4cos2x+cos4x)
= 1/8(3 - 4 + 8sin^2 x + 1 - 8sin^2 x + 8 sin^4 x)
= 1/8(8sin^4 x)
= sin^4 x
To prove the given identities using power reducing identities, we need to rewrite the left side of each equation in terms of trigonometric functions with lesser powers. Let's start with the first identity:
Identity 1: sin^4x = (1/8)(3 - 4cos2x + cos4x)
Using the power reducing identity for sine, sin^2x = (1/2)(1 - cos2x), we can rewrite sin^4x as (sin^2x)^2:
sin^4x = (sin^2x)^2 = ((1/2)(1 - cos2x))^2 = (1/4)(1 - cos2x)^2
Expanding (1 - cos2x)^2 using the binomial formula, we get:
sin^4x = (1/4)(1 - 2cos2x + cos^2(2x))
Now, using the power reducing identity for cosine, cos^2x = (1/2)(1 + cos2x), we can substitute it into the above equation:
sin^4x = (1/4)(1 - 2cos2x + (1/2)(1 + cos2x))
Expanding and simplifying, we get:
sin^4x = (1/4)(1 - 2cos2x + 1/2 + cos2x/2)
= (1/4)(3/2 - 3cos2x/2)
= (1/8)(3 - 3cos2x)
= (1/8)(3 - 4cos2x + cos4x)
Thus, we have proved Identity 1 using power reducing identities.
Moving on to the second identity:
Identity 2: cos^3x = (1/2cosx)(1 + cos2x)
Using the power reducing identity for cosine, cos^2x = (1/2)(1 + cos2x), we can rewrite cos^3x as cosx(cos^2x):
cos^3x = cosx(cos^2x) = cosx[(1/2)(1 + cos2x)]
Simplifying, we get:
cos^3x = (1/2cosx)(1 + cos2x)
Therefore, Identity 2 is proven using power reducing identities.
To summarize:
Identity 1: sin^4x = (1/8)(3 - 4cos2x + cos4x) [used power reducing identities for sine and cosine]
Identity 2: cos^3x = (1/2cosx)(1 + cos2x) [used power reducing identity for cosine]
To prove the identities, we will make use of the following power reducing identities:
1. sin²x = (1 - cos2x)/2
2. cos²x = (1 + cos2x)/2
Let's start with the first identity:
sin⁴x = 1/8(3 - 4cos2x + cos4x)
Using the power reducing identities, we can rewrite sin⁴x as:
(sin²x)² = 1/8(3 - 4cos2x + cos4x)
Now, substitute the values from the power reducing identities:
[(1 - cos2x)/2]² = 1/8(3 - 4cos2x + cos4x)
Squaring the expression on the left side:
(1 - cos2x)²/4 = 1/8(3 - 4cos2x + cos4x)
Now, multiply through by 4 to eliminate the denominators:
(1 - cos2x)² = 1/2(3 - 4cos2x + cos4x)
Expanding the left side:
1 - 2cos2x + cos²2x = 1/2(3 - 4cos2x + cos4x)
Since cos²x = (1 + cos2x)/2, we can rewrite the equation as:
1 - 2cos2x + (1 + cos2x)/2 = 1/2(3 - 4cos2x + cos4x)
Multiplying through by 2 to eliminate the denominator:
2 - 4cos2x + 1 + cos2x = 3 - 4cos2x + cos4x
Combining like terms:
3 - 3cos2x = 3 - 3cos2x + cos4x
Canceling out like terms:
3 - 3cos2x = 3 - 3cos2x + cos4x
The left side of the equation is equal to the right side, thus proving the identity.
Now, let's move on to the second identity:
cos³x = (1/2cosx)(1 + cos2x)
Starting with the right side of the equation:
(1/2cosx)(1 + cos2x)
Using the power reducing identity cos²x = (1 + cos2x)/2, we can simplify the expression:
(1/2cosx)(1 + (1 + cos2x)/2)
Combining like terms:
(1/2cosx)(2 + 1 + cos2x)
Simplifying further:
(1/2cosx)(3 + cos2x)
Now, distribute the (1/2cosx) to both terms:
(1/2cosx)(3) + (1/2cosx)(cos2x)
Simplifying each term separately:
3/2 + (1/2cosx)(cos2x)
Using the identity cosx·cos2x = cos3x/2, we can substitute the term:
3/2 + (1/2cosx)(cos3x/2)
Combining the terms:
3/2 + cos³x/4
Now, we have the left side of the equation:
cos³x = 3/2 + cos³x/4
Multiply through by 4 to eliminate the denominator:
4cos³x = 6 + cos³x
Subtract cos³x from both sides:
3cos³x = 6
Divide both sides by 3:
cos³x = 2
This completes the proof of the identity.
Hence, we have successfully proven the given identities using power reducing identities.