Question

The exact Value of the sine, cosine, and tangent about of the angle

Answer 1

$-\frac{11\pi}{12}=\frac{\pi}{4}-\frac{7\pi}{6}$

Answer:

Below are the exact values of sine, cosine, and tangent of the given angle -11π/12.

$sin(-\frac{11\pi}{12})=\frac{-\sqrt{6}+\sqrt{2}}{4}$$cos(-\frac{11\pi}{12})=\frac{-\sqrt{6}-\sqrt{2}}{4}$$tan(-\frac{11\pi}{12})=\frac{sin(-\frac{11\pi}{12})}{cos(-\frac{11\pi}{12})}=2-\sqrt{3}$

Explanation:

We can use the following trigonometric identities to solve the exact value of the given angle.

$\begin{gathered} sin(x+y)=sin\text{ }x\text{ }cos\text{ }y+cos\text{ }x\text{ }sin\text{ }y \\ cos(x+y)=cos\text{ }x\text{ }cos\text{ }y-sin\text{ }x\text{ }sin\text{ }y \end{gathered}$

For sine function, we have:

$\begin{gathered} sin(-\frac{11\pi}{12})=sin(\frac{\pi}{4}-\frac{7\pi}{6}) \\ sin(\frac{\pi}{4}-\frac{7\pi}{6})=sin\text{ }\frac{\pi}{4}cos(-\frac{7\pi}{6})+cos\text{ }\frac{\pi}{4}sin(-\frac{7\pi}{6}) \end{gathered}$

Simplify.

$\begin{gathered} sin(\frac{\pi}{4}-\frac{7\pi}{6})=\frac{\sqrt{2}}{2}(-\frac{\sqrt{3}}{2})+\frac{\sqrt{2}}{2}(\frac{1}{2}) \\ sin(\frac{\pi}{4}-\frac{7\pi}{6})=\frac{-\sqrt{6}}{4}+\frac{\sqrt{2}}{4} \\ sin(\frac{\pi}{4}-\frac{7\pi}{6})=\frac{-\sqrt{6}+\sqrt{2}}{4} \\ sin(-\frac{11\pi}{12})=\frac{-\sqrt{6}+\sqrt{2}}{4} \end{gathered}$

For cosine function, we have:

$\begin{gathered} cos(-\frac{11\pi}{12})=cos(\frac{\pi}{4}-\frac{7\pi}{6}) \\ cos(\frac{\pi}{4}-\frac{7\pi}{6})=cos\frac{\pi}{4}cos(-\frac{7\pi}{6})-sin\frac{\pi}{4}sin(-\frac{7\pi}{6}) \end{gathered}$

Simplify.

$\begin{gathered} cos(\frac{\pi}{4}-\frac{7\pi}{6})=\frac{\sqrt{2}}{2}(-\frac{\sqrt{3}}{2})-(\frac{\sqrt{2}}{2})(\frac{1}{2}) \\ cos(\frac{\pi}{4}-\frac{7\pi}{6})=\frac{-\sqrt{6}}{4}-\frac{\sqrt{2}}{4} \\ cos(\frac{\pi}{4}-\frac{7\pi}{6})=\frac{-\sqrt{6}-\sqrt{2}}{4} \\ cos(-\frac{11\pi}{12})=\frac{-\sqrt{6}-\sqrt{2}}{4} \end{gathered}$

Lastly, for tangent function, it is the ratio between sine and cosine function.

$tan(-\frac{11\pi}{12})=\frac{sin(-\frac{11\pi}{12})}{cos(-\frac{11\pi}{12})}$

Applying the division rule, we get the reciprocal of the denominator and multiply it to the numerator. The equation becomes:

$tan(-\frac{11\pi}{12})=sin(-\frac{11\pi}{12})\times\frac{1}{cos(-\frac{11\pi}{12})}$

Now, let's replace the sine and cosine value that we have calculated above.

$tan(-\frac{11\pi}{12})=\frac{-\sqrt{6}+\sqrt{2}}{4}\times\frac{4}{-\sqrt{6}-\sqrt{2}}$

Since 4 is a common factor on both numerator and denominator, we can cancel it out. The equation then becomes,

$tan(-\frac{11\pi}{12})=\frac{-\sqrt{6}+\sqrt{2}}{-\sqrt{6}-\sqrt{2}}$

To further simplify the function, let's remove the radicals in the denominator by rationalization.

$\begin{gathered} tan(-\frac{11\pi}{12})=\frac{-\sqrt{6}+\sqrt{2}}{-\sqrt{6}-\sqrt{2}}\times\frac{-\sqrt{6}+\sqrt{2}}{-\sqrt{6}+\sqrt{2}} \\ tan(-\frac{11\pi}{12})=\frac{6-\sqrt{12}-\sqrt{12}+2}{6-\sqrt{12}+\sqrt{12}-2} \end{gathered}$$\begin{gathered} tan(-\frac{11\pi}{12})=\frac{8-2\sqrt{12}}{4} \\ tan(-\frac{11\pi}{12})=\frac{8-2\sqrt{4\times3}}{4} \\ tan(-\frac{11\pi}{12})=\frac{8-(2\times2\sqrt{3})}{4} \\ tan(-\frac{11\pi}{12})=\frac{8-4\sqrt{3}}{4} \\ Factor\text{ }4\text{ }in\text{ }the\text{ }numerator. \\ tan(-\frac{11\pi}{12})=\frac{4(2-\sqrt{3})}{4} \\ Cancel\text{ }4. \\ tan(-\frac{11\pi}{12})=2-\sqrt{3} \end{gathered}$