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Find the exact value for all six trig functions of the angle

Find The Exact Value For All Six Trig Functions Of The Angle class=

Sagot :

We are asked to determine the trigonometric functions for:

[tex]\theta=\frac{23\pi}{6}[/tex]

To determine the trigonometric functions we need to determine the equivalent angle that is between 0 and 2pi. To do that we will subtract 2pi from the given angle:

[tex]\frac{23\pi}{6}-2\pi=\frac{11\pi}{6}[/tex]

The equivalent angle is 11pi/6. In the unit circle this angle is:

The end-point of this angle is:

[tex](\frac{\sqrt{3}}{2},-\frac{1}{2})[/tex]

Since the x-component of the end-point is the cosine, we have:

[tex]cos(\frac{23\pi}{6})=\frac{\sqrt{3}}{2}[/tex]

The y-coordinate is the sine, therefore:

[tex]sin(\frac{23\pi}{6})=-\frac{1}{2}[/tex]

To determine the tangent we use the following relationship:

[tex]tanx=\frac{sinx}{cosx}[/tex]

Substituting we get:

[tex]tan(\frac{23\pi}{6})=\frac{-\frac{1}{2}}{\frac{\sqrt{3}}{2}}[/tex]

Simplifying we get:

[tex]\begin{gathered} tan(\frac{23\pi}{6})=\frac{-1}{\sqrt{3}} \\ \end{gathered}[/tex]

To determine the secant we use the following relationship:

[tex]secx=\frac{1}{cosx}[/tex]

Substituting we get:

[tex]sec(\frac{23\pi}{6})=\frac{1}{\frac{\sqrt{3}}{2}}[/tex]

Simplifying we get:

[tex]sec(\frac{23\pi}{6})=\frac{2}{\sqrt{3}}[/tex]

To determine the cosecant we use the following relationship:

[tex]cscx=\frac{1}{sinx}[/tex]

Substituting we get:

[tex]csc(\frac{23\pi}{6})=\frac{1}{-\frac{1}{2}}[/tex]

Simplifying we get:

[tex]csc(\frac{23\pi}{6})=-2[/tex]

Finally, for the cotangent we use:

[tex]ctgx=\frac{1}{tanx}[/tex]

Substituting we get:

[tex]ctgx=\frac{1}{\frac{-1}{\sqrt{3}}}[/tex]

Simplifying:

[tex]ctg(\frac{23\pi}{6})=-\sqrt{3}[/tex]

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