Sagot :
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]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
