Question

all graphs are -8,8,2 by -8,8,2.
use the graph to make a conjecture as to what might be an identity. then verify your conjecture algebraically. choose the correct answer below.
a.
2\frac{\cot \frac{x}{2}-\tan \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{\cos \frac{x}{2}+\sin \frac{x}{2}}=\cot x
b.
2\frac{\cot \frac{x}{2}-\tan \frac{x}{2}}{\cos \frac{x}{2}-\sin \frac{x}{2}}=\frac{\cos \frac{x}{2}-\sin \frac{x}{2}}{1}=\sec \frac{x}{2}
c.
2\frac{\cot \frac{x}{2}-\tan \frac{x}{2}}{2\sin \frac{x}{2}\cos \frac{x}{2}}=\frac{1-\cos x}{\sin \frac{x}{2}}=\tan \frac{x}{2}
choose the correct graph below.
oa.
ob.
oc.
od.
\frac{\cot \frac{x}{2}-\tan \frac{x}{2}}{2}

Explanation:

Step1: Use trigonometric identities

We know that \(\cot\frac{x}{2}=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}\) and \(\tan\frac{x}{2}=\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}\). Then \(\cot\frac{x}{2}-\tan\frac{x}{2}=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}-\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\frac{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}{\sin\frac{x}{2}\cos\frac{x}{2}}\).
By the double - angle formulas: \(\cos^{2}\alpha-\sin^{2}\alpha = \cos2\alpha\) and \(2\sin\alpha\cos\alpha=\sin2\alpha\). So \(\cot\frac{x}{2}-\tan\frac{x}{2}=\frac{\cos x}{\frac{1}{2}\sin x}\) (since \(\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}=\cos x\) and \(2\sin\frac{x}{2}\cos\frac{x}{2}=\sin x\)).

Step2: Simplify the expression

\(\frac{\cos x}{\frac{1}{2}\sin x}=\frac{2\cos x}{\sin x}=2\cot x\).

Answer:

A. \(\cot\frac{x}{2}-\tan\frac{x}{2}=\frac{\cos\frac{x}{2}}{\sin\frac{x}{2}}-\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}}=\frac{\cos^{2}\frac{x}{2}-\sin^{2}\frac{x}{2}}{\sin\frac{x}{2}\cos\frac{x}{2}}=\frac{\cos x}{\frac{1}{2}\sin x}=2\cot x\)