201k views
5 votes
I have a trigonometric question I will upload a photo

I have a trigonometric question I will upload a photo-example-1
User Pmko
by
4.7k points

1 Answer

5 votes

The secant can be calculated as the inverse of the cosine, so we have:


\sec (-(7\pi)/(3))=(1)/(\cos (-(7\pi)/(3)))=(1)/(0.5)=2

The cosecant can be calculated as the inverse of the sine, so we have:


\csc (-(7\pi)/(3))=(1)/(\sin(-(7\pi)/(3)))=\frac{1}{\frac{\sqrt[]{3}}{2}}=\frac{2}{\sqrt[]{3}}=\frac{2\sqrt[]{3}}{3}=1.1547

The tangent can be calculated as the sine over the cosine, so we have:


\tan (-(7\pi)/(3))=(\sin(-(7\pi)/(3)))/(\cos(-(7\pi)/(3)))=\frac{\frac{\sqrt[]{3}}{2}}{0.5}=\sqrt[]{3}=1.732

The cotangent can be calculated as the inverse of the tangent, so we have:


\cot (-(7\pi)/(3))=(1)/(\tan(-(7\pi)/(3)))=\frac{1}{\sqrt[]{3}}=\frac{\sqrt[]{3}}{3}=0.57735

User Yohani
by
5.2k points