Question

match each expression in column i with its value in column ii.
for each expression in column i, type the letter that matches its value in colum
$2\sin(30^{\circ})\cos(30^{\circ})$:
$1 - 2\sin^{2}\frac{\pi}{8}$:
$\frac{2\tan\frac{\pi}{12}}{1 - \tan^{2}\frac{\pi}{12}}$:
$2\sin\frac{\pi}{12}\cos\frac{\pi}{12}$:
$4\sin\frac{\pi}{3}\cos\frac{\pi}{3}$:
$\frac{2\tan\frac{\pi}{3}}{1 - \tan^{2}\frac{\pi}{3}}$:
a. $\sqrt{3}$
b. $\frac{1}{2}$
c. $\frac{\sqrt{2}}{2}$
d. $-\sqrt{3}$
e. $\frac{\sqrt{3}}{2}$
f. $\frac{\sqrt{3}}{3}$

Explanation:

Step1: Use double - angle formula for \(2\sin(30^{\circ})\cos(30^{\circ})\)

The double - angle formula for sine is \(\sin(2\alpha)=2\sin\alpha\cos\alpha\).
When \(\alpha = 30^{\circ}\), then \(2\sin(30^{\circ})\cos(30^{\circ})=\sin(60^{\circ})\).
Since \(\sin(60^{\circ})=\frac{\sqrt{3}}{2}\), so \(2\sin(30^{\circ})\cos(30^{\circ})\) matches with \(E\).

Step2: Use double - angle formula for \(1 - 2\sin^{2}\frac{\pi}{8}\)

The double - angle formula for cosine is \(\cos(2\alpha)=1 - 2\sin^{2}\alpha\).
When \(\alpha=\frac{\pi}{8}\), then \(1 - 2\sin^{2}\frac{\pi}{8}=\cos(\frac{\pi}{4})\).
Since \(\cos(\frac{\pi}{4})=\frac{\sqrt{2}}{2}\), so \(1 - 2\sin^{2}\frac{\pi}{8}\) matches with \(C\).

Step3: Use double - angle formula for \(\frac{2\tan\frac{\pi}{12}}{1-\tan^{2}\frac{\pi}{12}}\)

The double - angle formula for tangent is \(\tan(2\alpha)=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}\).
When \(\alpha=\frac{\pi}{12}\), then \(\frac{2\tan\frac{\pi}{12}}{1-\tan^{2}\frac{\pi}{12}}=\tan(\frac{\pi}{6})\).
Since \(\tan(\frac{\pi}{6})=\frac{\sqrt{3}}{3}\), so \(\frac{2\tan\frac{\pi}{12}}{1-\tan^{2}\frac{\pi}{12}}\) matches with \(F\).

Step4: Use double - angle formula for \(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\)

Using the double - angle formula \(\sin(2\alpha)=2\sin\alpha\cos\alpha\).
When \(\alpha=\frac{\pi}{12}\), then \(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}=\sin(\frac{\pi}{6})\).
Since \(\sin(\frac{\pi}{6})=\frac{1}{2}\), so \(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}\) matches with \(B\).

Step5: Use double - angle formula for \(4\sin\frac{\pi}{3}\cos\frac{\pi}{3}\)

First, rewrite \(4\sin\frac{\pi}{3}\cos\frac{\pi}{3}=2\times(2\sin\frac{\pi}{3}\cos\frac{\pi}{3})\).
Using the double - angle formula \(\sin(2\alpha)=2\sin\alpha\cos\alpha\) with \(\alpha=\frac{\pi}{3}\), we get \(2\sin(\frac{2\pi}{3})\).
Since \(\sin(\frac{2\pi}{3})=\frac{\sqrt{3}}{2}\), then \(2\sin(\frac{2\pi}{3})=\sqrt{3}\), so \(4\sin\frac{\pi}{3}\cos\frac{\pi}{3}\) matches with \(A\).

Step6: Use double - angle formula for \(\frac{2\tan\frac{\pi}{3}}{1-\tan^{2}\frac{\pi}{3}}\)

Using the double - angle formula \(\tan(2\alpha)=\frac{2\tan\alpha}{1 - \tan^{2}\alpha}\).
When \(\alpha=\frac{\pi}{3}\), then \(\frac{2\tan\frac{\pi}{3}}{1-\tan^{2}\frac{\pi}{3}}=\tan(\frac{2\pi}{3})\).
Since \(\tan(\frac{2\pi}{3})=-\sqrt{3}\), so \(\frac{2\tan\frac{\pi}{3}}{1-\tan^{2}\frac{\pi}{3}}\) matches with \(D\).

Answer:

\(2\sin(30^{\circ})\cos(30^{\circ}):E\)
\(1 - 2\sin^{2}\frac{\pi}{8}:C\)
\(\frac{2\tan\frac{\pi}{12}}{1-\tan^{2}\frac{\pi}{12}}:F\)
\(2\sin\frac{\pi}{12}\cos\frac{\pi}{12}:B\)
\(4\sin\frac{\pi}{3}\cos\frac{\pi}{3}:A\)
\(\frac{2\tan\frac{\pi}{3}}{1-\tan^{2}\frac{\pi}{3}}:D\)