In b we need to find:
[tex]\sec 18\pi[/tex]It's important to recal that the secant is equal to:
[tex]\sec 18\pi=\frac{1}{\cos18\pi}[/tex]Another important property that will be useful is:
[tex]\cos x=\cos (x+2\pi m)[/tex]Where m is any integer. Let's see if we can write 18*pi using this. We can take x=0 so we have:
[tex]\begin{gathered} 18\pi=x+2\pi m=2\pi m \\ 18\pi=2\pi m \end{gathered}[/tex]If we divide both sides by 2*pi:
[tex]\begin{gathered} \frac{18\pi}{2\pi}=\frac{2\pi m}{2\pi} \\ 9=m \end{gathered}[/tex]Since m is an integer then we can assure that:
[tex]\cos 18\pi=\cos (0+2\pi\cdot9)=\cos 0=1[/tex]Then the secant is given by:
[tex]\sec 18\pi=\frac{1}{\cos18\pi}=\frac{1}{\cos 0}=1[/tex]So the answer to b is 1.
In c we need to find:
[tex]\sin (\frac{7\pi}{6})\tan (\frac{8\pi}{3})[/tex]Here we can use the following properties in order to write those angles as angles of the first quadrant:
[tex]\begin{gathered} \sin (x)=-\sin (x-\pi) \\ \tan (x)=\tan (x-m\pi)\text{ with }m\text{ being an integer} \end{gathered}[/tex]So we have:
[tex]\begin{gathered} \sin (\frac{7\pi}{6})=-\sin (\frac{7\pi}{6}-\pi)=-\sin (\frac{\pi}{6}) \\ \tan (\frac{8\pi}{3})=\tan (\frac{8\pi}{3}-3\pi)=\tan (-\frac{1}{3}\pi) \end{gathered}[/tex]If we convert these two angles from radians to degrees by multiplying 360° and dividing by 2*pi we have:
[tex]\begin{gathered} \frac{\pi}{6}\cdot\frac{360^{\circ}}{2\pi}=30^{\circ} \\ -\frac{1}{3}\pi\cdot\frac{360^{\circ}}{2\pi}=-60^{\circ} \end{gathered}[/tex]And remeber that:
[tex]\tan x=-\tan (-x)[/tex]So we get:
[tex]\begin{gathered} \sin (\frac{7\pi}{6})=-\sin (\frac{\pi}{6})=-\sin (30^{\circ}) \\ \tan (\frac{8\pi}{3})=\tan (-\frac{\pi}{3})=-\tan (\frac{\pi}{3})=-\tan (60^{\circ}) \end{gathered}[/tex]Then we can use a table of values:
Then:
[tex]\sin (\frac{7\pi}{6})\tan (\frac{8\pi}{3})=\sin (30^{\circ})\cdot\tan (60^{\circ})=\frac{1}{2}\cdot\sqrt[]{3}=\frac{\sqrt[]{3}}{2}[/tex]So the answer to c is (√3)/2.
In d we need to find:
[tex]\tan (\frac{\pi}{12})[/tex]In order to do this using the table we can use the following:
[tex]\begin{gathered} \tan x=\frac{\sin x}{\cos x} \\ \sin 2x=2\sin x\cos x \\ \cos 2x=\cos ^2x-\sin ^2x \\ \cos ^2x+\sin ^2x=1 \end{gathered}[/tex]So from the first one we have:
[tex]\tan (\frac{\pi}{12})=\frac{\sin (\frac{\pi}{12})}{\cos (\frac{\pi}{12})}[/tex]We convert pi/12 into degrees:
[tex]\frac{\pi}{12}\cdot\frac{360^{\circ}}{2\pi}=15^{\circ}[/tex]So we need to find the sine and cosine of 15°. We use the second equation:
[tex]\begin{gathered} \sin 30^{\circ}=\frac{1}{2}=\sin (2\cdot15^{\circ})=2\sin 15^{\circ}\cos 15^{\circ} \\ \sin 15^{\circ}\cos 15^{\circ}=\frac{1}{4} \end{gathered}[/tex]Then we use the third:
[tex]\begin{gathered} \cos (30^{\circ})=\frac{\sqrt[]{3}}{2}=\cos (2\cdot15^{\circ})=\cos ^215^{\circ}-\sin ^215^{\circ} \\ \frac{\sqrt[]{3}}{2}=\cos ^215^{\circ}-\sin ^215^{\circ} \end{gathered}[/tex]And from the fourth equation we get:
[tex]\begin{gathered} \cos ^215^{\circ}+\sin ^215^{\circ}=1 \\ \sin ^215^{\circ}=1-\cos ^215^{\circ} \end{gathered}[/tex]We can use this in the previous equation:
[tex]\begin{gathered} \frac{\sqrt[]{3}}{2}=\cos ^215^{\circ}-\sin ^215^{\circ}=\cos ^215^{\circ}-(1-\cos ^215^{\circ}) \\ \frac{\sqrt[]{3}}{2}=2\cos ^215^{\circ}-1 \\ \cos 15^{\circ}=\sqrt{\frac{1+\frac{\sqrt[]{3}}{2}}{2}} \\ \cos 15^{\circ}=\sqrt{\frac{1}{2}+\frac{\sqrt[]{3}}{4}} \end{gathered}[/tex]So we found the cosine. For the sine we use the expression with the sine and cosine multiplying:
[tex]\begin{gathered} \sin 15^{\circ}\cos 15^{\circ}=\frac{1}{4} \\ \sin 15^{\circ}\cdot\sqrt[]{\frac{1}{2}+\frac{\sqrt[]{3}}{4}}=\frac{1}{4} \\ \sin 15^{\circ}=\frac{1}{4\cdot\sqrt[]{\frac{1}{2}+\frac{\sqrt[]{3}}{4}}} \end{gathered}[/tex]Then the tangent is:
[tex]\tan (15^{\circ})=\frac{\sin(15^{\circ})}{\cos(15^{\circ})}=\frac{1}{4\cdot\sqrt[]{\frac{1}{2}+\frac{\sqrt[]{3}}{4}}}\cdot\frac{1}{\sqrt[]{\frac{1}{2}+\frac{\sqrt[]{3}}{4}}}=\frac{1}{4}\cdot\frac{1}{\frac{1}{2}+\frac{\sqrt[]{3}}{4}}[/tex][tex]\tan (15^{\circ})=\frac{1}{4}\cdot\frac{1}{\frac{1}{2}+\frac{\sqrt[]{3}}{4}}=\frac{1}{2+\sqrt[]{3}}[/tex]Then the answer to d is:
[tex]\frac{1}{2+\sqrt[]{3}}[/tex]
