Can somebody show I steps to simplify this [9(sin(t)^4 cos(t)^2 +cos(t)^4 sin(t)^2)]/[9(cos(t)^4 sin(t)^2 +sin(t)^4 cos(t)^2 )]^(3/2
well, you can clearly factor out a 9sin^2t cos^2t. That leaves
9sin^2t cos^2t [(sin(t)^2 +cos(t)^2)] / 27sin^3t cos^3t[(cos(t)^2+sin(t)^2)]^(3/2)
Think you can handle that?
To simplify the expression, let's first rewrite it in a more organized way:
[9(sin(t)^4 cos(t)^2 + cos(t)^4 sin(t)^2)] / [9(cos(t)^4 sin(t)^2 + sin(t)^4 cos(t)^2)]^(3/2)
Now, let's simplify the numerator and denominator separately before combining them:
1. Simplifying the numerator:
The numerator has a common factor of 9, so we can factor it out:
[9(sin(t)^4 cos(t)^2 + cos(t)^4 sin(t)^2)] = 9(sin(t)^2 cos(t)^2)(sin(t)^2 + cos(t)^2)
Since sin(t)^2 + cos(t)^2 = 1 (from the Pythagorean identity), we can simplify further:
9(sin(t)^2 cos(t)^2)(1) = 9(sin(t)^2 cos(t)^2)
2. Simplifying the denominator:
The denominator has a common factor of 9 as well, so we can factor it out:
[9(cos(t)^4 sin(t)^2 + sin(t)^4 cos(t)^2)] = 9(sin(t)^2 cos(t)^2)(sin(t)^2 + cos(t)^2)
Again, since sin(t)^2 + cos(t)^2 = 1, we can simplify further:
9(sin(t)^2 cos(t)^2)(1) = 9(sin(t)^2 cos(t)^2)
Now, let's substitute the simplified numerator and denominator back into the original expression:
[9(sin(t)^2 cos(t)^2)] / [9(sin(t)^2 cos(t)^2)]^(3/2)
Simplifying further, we use the property of exponents that states (a^m)^n = a^(m*n):
[9(sin(t)^2 cos(t)^2)] / [9^(3/2)(sin(t)^2 cos(t)^2)^(3/2)]
Since 9^(3/2) equals 27, and the exponent (3/2) can be considered as a square root, we have:
[9(sin(t)^2 cos(t)^2)] / [27(sin(t)^2 cos(t)^2)]
Now, we can cancel out the common factor of (sin(t)^2 cos(t)^2):
[9(sin(t)^2 cos(t)^2)] / [27(sin(t)^2 cos(t)^2)] = 1/3
So, the simplified expression is 1/3.