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test 6.3 double half angle
(8 points)
by using known trig identities, \\(\frac{\sin(2x)}{1+\cos(2x)}\\) can be written as
\\(\bigcirc\\) a. \\(\tan(2x)\\)
\\(\bigcirc\\) b. \\(\sec(x)\\)
\\(\bigcirc\\) c. \\(\tan(x)\\)
\\(\bigcirc\\) d. \\(\csc(2x)\\)
\\(\bigcirc\\) e. all of the above
\\(\bigcirc\\) f. none of the above

Explanation:

Step1: Recall double-angle identities

We know that \(\sin(2x) = 2\sin(x)\cos(x)\) and \(\cos(2x)=2\cos^{2}(x)-1\), so \(1 + \cos(2x)=2\cos^{2}(x)\).

Step2: Substitute identities into the expression

Substitute \(\sin(2x) = 2\sin(x)\cos(x)\) and \(1+\cos(2x)=2\cos^{2}(x)\) into \(\frac{\sin(2x)}{1 + \cos(2x)}\), we get \(\frac{2\sin(x)\cos(x)}{2\cos^{2}(x)}\).

Step3: Simplify the expression

Cancel out the common factors \(2\) and \(\cos(x)\) (assuming \(\cos(x)
eq0\)), we have \(\frac{\sin(x)}{\cos(x)}=\tan(x)\).

Answer:

C. \(\tan(x)\)