Libby Owens

2022-07-27

t is a real number and P=(x,y) is the point on the unit circlethar corresponds to t. Find the exact values of the six trigonometric functions of t.

$(\frac{1}{2},\frac{-\sqrt{3}}{2})$

$(\frac{1}{2},\frac{-\sqrt{3}}{2})$

phinny5608tt

Beginner2022-07-28Added 17 answers

t is a real number and P=(x,y) isthe point on the unit circle thar corresponds to t. Find the exact values of the six trigonometric functions of t.

according to point we are on the fourth quadrant. As we know on the fourth quadrant (x,y) x positive and y is the negative sign.

$\mathrm{sin}(x)=-\frac{\sqrt{3}}{2}$

$\mathrm{cos}(x)=\frac{1}{2}$

$\mathrm{tan}(x)=-\sqrt{3}$

$\mathrm{cot}(x)=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$

$\mathrm{sec}(x)=1/\mathrm{cos}(x)=\frac{1}{\frac{1}{2}}=2$

$cosec(x)=1/\mathrm{sin}(x)=-\frac{1}{\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}$

according to point we are on the fourth quadrant. As we know on the fourth quadrant (x,y) x positive and y is the negative sign.

$\mathrm{sin}(x)=-\frac{\sqrt{3}}{2}$

$\mathrm{cos}(x)=\frac{1}{2}$

$\mathrm{tan}(x)=-\sqrt{3}$

$\mathrm{cot}(x)=-\frac{1}{\sqrt{3}}=-\frac{\sqrt{3}}{3}$

$\mathrm{sec}(x)=1/\mathrm{cos}(x)=\frac{1}{\frac{1}{2}}=2$

$cosec(x)=1/\mathrm{sin}(x)=-\frac{1}{\frac{\sqrt{3}}{2}}=-\frac{2}{\sqrt{3}}$

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