Question

match each expression with the correct value
fill in each blank with the correct choice
$\cos(75^{\circ}) =$
$\tan(15^{\circ}) =$
$\sin\frac{3\pi}{8} =$
$\tan(-\frac{\pi}{8})$
$\tan\frac{3\pi}{8} =$
$\cos(675^{\circ})$
$\frac{\sqrt{2 - \sqrt{3}}}{2}$
$2 - \sqrt{3}$
$1 - \sqrt{2}$
$\frac{\sqrt{2 - \sqrt{2}}}{2}$
$1 + \sqrt{2}$
$\frac{\sqrt{2 + \sqrt{2}}}{2}$

Explanation:

Step1: Use the cosine of sum formula

$\cos(A + B)=\cos A\cos B-\sin A\sin B$. For $\cos(75^{\circ})=\cos(45^{\circ}+ 30^{\circ})$, we have $\cos45^{\circ}=\frac{\sqrt{2}}{2}$, $\sin45^{\circ}=\frac{\sqrt{2}}{2}$, $\cos30^{\circ}=\frac{\sqrt{3}}{2}$, $\sin30^{\circ}=\frac{1}{2}$. Then $\cos(75^{\circ})=\frac{\sqrt{2}}{2}\times\frac{\sqrt{3}}{2}-\frac{\sqrt{2}}{2}\times\frac{1}{2}=\frac{\sqrt{6}-\sqrt{2}}{4}=\frac{\sqrt{2-\sqrt{3}}}{2}$ (after rationalizing).

Step2: Use the tangent of difference formula

$\tan(A - B)=\frac{\tan A-\tan B}{1 + \tan A\tan B}$. For $\tan(15^{\circ})=\tan(45^{\circ}-30^{\circ})$, $\tan45^{\circ}=1$, $\tan30^{\circ}=\frac{\sqrt{3}}{3}$. Then $\tan(15^{\circ})=\frac{1-\frac{\sqrt{3}}{3}}{1 + 1\times\frac{\sqrt{3}}{3}}=\frac{3-\sqrt{3}}{3 + \sqrt{3}}=2-\sqrt{3}$ (after rationalizing).

Step3: Use the sine of half - angle formula

$\sin\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{2}}$. For $\sin\frac{3\pi}{8}$, let $\alpha=\frac{3\pi}{4}$, then $\sin\frac{3\pi}{8}=\sqrt{\frac{1-\cos\frac{3\pi}{4}}{2}}=\sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}=\frac{\sqrt{2 + \sqrt{2}}}{2}$.

Step4: Use the tangent of negative angle formula and half - angle formula

$\tan(-x)=-\tan x$ and $\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}$. For $\tan(-\frac{\pi}{8})=-\tan\frac{\pi}{8}$, let $\alpha=\frac{\pi}{4}$, then $\tan\frac{\pi}{8}=\frac{1-\cos\frac{\pi}{4}}{\sin\frac{\pi}{4}}=\sqrt{2}-1$, so $\tan(-\frac{\pi}{8})=1 - \sqrt{2}$.

Step5: Use the tangent of half - angle formula

$\tan\frac{\alpha}{2}=\frac{1-\cos\alpha}{\sin\alpha}$. For $\tan\frac{3\pi}{8}$, let $\alpha=\frac{3\pi}{4}$, then $\tan\frac{3\pi}{8}=\frac{1-\cos\frac{3\pi}{4}}{\sin\frac{3\pi}{4}}=\frac{1+\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1+\sqrt{2}$.

Step6: Use the cosine of half - angle formula

$\cos\frac{\alpha}{2}=\sqrt{\frac{1+\cos\alpha}{2}}$. For $\cos(67.5^{\circ})=\cos\frac{135^{\circ}}{2}$, let $\alpha = 135^{\circ}$, then $\cos(67.5^{\circ})=\sqrt{\frac{1+\cos135^{\circ}}{2}}=\frac{\sqrt{2-\sqrt{2}}}{2}$.

Answer:

$\cos(75^{\circ})=\frac{\sqrt{2-\sqrt{3}}}{2}$; $\tan(15^{\circ})=2-\sqrt{3}$; $\sin\frac{3\pi}{8}=\frac{\sqrt{2+\sqrt{2}}}{2}$; $\tan(-\frac{\pi}{8})=1-\sqrt{2}$; $\tan\frac{3\pi}{8}=1+\sqrt{2}$; $\cos(67.5^{\circ})=\frac{\sqrt{2-\sqrt{2}}}{2}$