Factor 1-sin^3x
let a = sin x
then we have
1-a^3
which is
(1-a)(1+a+a^2)
so
(1-sin x)(1 + sin x + sin^2 x)
To factor the expression 1-sin^3x, we need to recognize that it is a difference of cubes.
The expression 1-sin^3x can be written as (1-sinx)(1+sin^2x).
Next, we can simplify further using the identity sin^2x = 1-cos^2x:
1+sin^2x = 1+(1-cos^2x) = 1+1-cos^2x = 2-cos^2x.
Therefore, the factored form of the expression 1-sin^3x is (1-sinx)(2-cos^2x).
To factor the expression 1 - sin^3(x), we can use the identity for the cube of a difference:
a^3 - b^3 = (a - b)(a^2 + ab + b^2),
where a = 1 and b = sin(x).
Applying this identity, we have:
1 - sin^3(x) = (1 - sin(x))(1 + sin(x) + sin^2(x)).
Now, let's simplify further. The expression (1 - sin(x)) cannot be simplified any further.
For the expression (1 + sin(x) + sin^2(x)), we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to rewrite sin^2(x) as 1 - cos^2(x):
1 + sin(x) + sin^2(x) = 1 + sin(x) + (1 - cos^2(x)).
Combining like terms, we have:
1 + sin(x) + 1 - cos^2(x) = 2 - cos^2(x) + sin(x).
Therefore, the factored form of 1 - sin^3(x) is:
(1 - sin(x))(2 - cos^2(x) + sin(x)).