Question

Hi, can you help me to evaluate (if possible) thesix trigonometric functions of the real number.Please.

Answer 1

Okay, here we have this:

Considering the provided angle, we are going to evaluate the trigonometric functions, so we obtain the following:

Sine:

`\begin{gathered} \sin (-(2\pi)/(3)) \\ =-\sin ((2\pi)/(3)) \\ =-\cos \mleft((\pi)/(2)-(2\pi)/(3)\mright) \\ =-\cos \mleft(-(\pi)/(6)\mright) \\ =-\cos \mleft((\pi)/(6)\mright) \\ =-(√(3))/(2) \end{gathered}`

Cos:

`\begin{gathered} cos\mleft(-(2\pi)/(3)\mright) \\ =\cos \mleft((2\pi)/(3)\mright) \\ =\sin \mleft((\pi)/(2)-(2\pi)/(3)\mright) \\ =\sin \mleft(-(\pi)/(6)\mright) \\ =-\sin \mleft((\pi)/(6)\mright) \\ =-(1)/(2) \end{gathered}`

Tan:

`\begin{gathered} tan\mleft(-(2\pi\:)/(3)\mright) \\ =(\sin (-(2\pi\: )/(3)))/(\cos (-(2\pi\: )/(3))) \\ =\frac{-\frac{\sqrt[]{3}}{2}}{-(1)/(2)} \\ =\sqrt[]{3} \end{gathered}`

Csc:

`\begin{gathered} \csc \mleft(-(2\pi)/(3)\mright) \\ =(1)/(\sin\left(-(2\pi)/(3)\right)) \\ =-(1)/((√(3))/(2)) \\ =-(2√(3))/(3) \end{gathered}`

Sec:

`\begin{gathered} \sec \mleft(-(2\pi)/(3)\mright) \\ =(1)/(\cos\left(-(2\pi)/(3)\right)) \\ =(1)/(-(1)/(2)) \\ =-2 \end{gathered}`

Cot:

`\begin{gathered} \cot \mleft(-(2\pi)/(3)\mright) \\ =(1)/(\tan (-(2\pi)/(3))) \\ =\frac{1}{\sqrt[]{3}} \\ =(√(3))/(3) \end{gathered}`