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test 6.3 double half angle power r
(8 points)
use identities to simplify each expression.
\\( \frac{\sin^3(-x)}{\sin^3(x) - \sin(x)} = \square \\)
\\( 1 - \frac{1}{\cos^2(x)} = \square \\)
\\( \frac{\cos(x)\sin^2(x) + \cos^3(x)}{\sec(x)} = \square \\)
note: you can earn partial credit on this problem.
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Explanation:

Step1: Simplify \(\frac{\sin^{3}(-x)}{\sin^{3}(x)-\sin(x)}\)

Recall that \(\sin(-x)=-\sin(x)\), so \(\sin^{3}(-x)=(-\sin(x))^{3}=-\sin^{3}(x)\).
Factor the denominator: \(\sin^{3}(x)-\sin(x)=\sin(x)(\sin^{2}(x)-1)=-\sin(x)(1 - \sin^{2}(x))=-\sin(x)\cos^{2}(x)\) (using \(\sin^{2}(x)+\cos^{2}(x)=1\)).
Now the expression becomes \(\frac{-\sin^{3}(x)}{-\sin(x)\cos^{2}(x)}=\frac{\sin^{2}(x)}{\cos^{2}(x)}=\tan^{2}(x)\) (using \(\tan(x)=\frac{\sin(x)}{\cos(x)}\)).

Step2: Simplify \(1-\frac{1}{\cos^{2}(x)}\)

Combine the terms: \(1-\frac{1}{\cos^{2}(x)}=\frac{\cos^{2}(x)-1}{\cos^{2}(x)}\).
Using \(\sin^{2}(x)+\cos^{2}(x)=1\), we have \(\cos^{2}(x)-1 = -\sin^{2}(x)\).
So the expression is \(\frac{-\sin^{2}(x)}{\cos^{2}(x)}=-\tan^{2}(x)\).

Step3: Simplify \(\frac{\cos(x)\sin^{2}(x)+\cos^{3}(x)}{\sec(x)}\)

Factor the numerator: \(\cos(x)\sin^{2}(x)+\cos^{3}(x)=\cos(x)(\sin^{2}(x)+\cos^{2}(x))\).
Since \(\sin^{2}(x)+\cos^{2}(x)=1\), the numerator becomes \(\cos(x)\times1=\cos(x)\).
Recall that \(\sec(x)=\frac{1}{\cos(x)}\), so dividing by \(\sec(x)\) is multiplying by \(\cos(x)\).
Thus, \(\cos(x)\times\cos(x)=\cos^{2}(x)\).

Answer:

\(\frac{\sin^{3}(-x)}{\sin^{3}(x)-\sin(x)}=\boxed{\tan^{2}(x)}\)
\(1 - \frac{1}{\cos^{2}(x)}=\boxed{-\tan^{2}(x)}\)
\(\frac{\cos(x)\sin^{2}(x)+\cos^{3}(x)}{\sec(x)}=\boxed{\cos^{2}(x)}\)