match each expression in column i with its value in column ii.\nfor each expression in column i, type the…

match each expression in column i with its value in column ii.\nfor each expression in column i, type the letter that matches its value in colum\n$2\\sin(30^{\\circ})\\cos(30^{\\circ})$: \n$1 - 2\\sin^{2}\\frac{\\pi}{8}$: \n$\\frac{2\\tan\\frac{\\pi}{12}}{1 - \\tan^{2}\\frac{\\pi}{12}}$: \n$2\\sin\\frac{\\pi}{12}\\cos\\frac{\\pi}{12}$: \n$4\\sin\\frac{\\pi}{3}\\cos\\frac{\\pi}{3}$: \n$\\frac{2\\tan\\frac{\\pi}{3}}{1 - \\tan^{2}\\frac{\\pi}{3}}$: \na. $\\sqrt{3}$\nb. $\\frac{1}{2}$\nc. $\\frac{\\sqrt{2}}{2}$\nd. $-\\sqrt{3}$\ne. $\\frac{\\sqrt{3}}{2}$\nf. $\\frac{\\sqrt{3}}{3}$

match each expression in column i with its value in column ii.\nfor each expression in column i, type the letter that matches its value in colum\n$2\\sin(30^{\\circ})\\cos(30^{\\circ})$: \n$1 - 2\\sin^{2}\\frac{\\pi}{8}$: \n$\\frac{2\\tan\\frac{\\pi}{12}}{1 - \\tan^{2}\\frac{\\pi}{12}}$: \n$2\\sin\\frac{\\pi}{12}\\cos\\frac{\\pi}{12}$: \n$4\\sin\\frac{\\pi}{3}\\cos\\frac{\\pi}{3}$: \n$\\frac{2\\tan\\frac{\\pi}{3}}{1 - \\tan^{2}\\frac{\\pi}{3}}$: \na. $\\sqrt{3}$\nb. $\\frac{1}{2}$\nc. $\\frac{\\sqrt{2}}{2}$\nd. $-\\sqrt{3}$\ne. $\\frac{\\sqrt{3}}{2}$\nf. $\\frac{\\sqrt{3}}{3}$

Answer

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)