Question

Trigonometric State the three basic trigonometric ratios for the following

Answer 1

Given

`\beta=-(7\pi)/(2)`

To find the three basic trigonometric ratios.

Explanation;

It is given that,

`\beta=-(7\pi)/(2)`

That implies,

`\begin{gathered} \sin\beta=sin(-(7\pi)/(2)) \\ =-\sin((7\pi)/(2)) \\ =-sin((6\pi)/(2)+(\pi)/(2)) \\ =-\sin(3\pi+(\pi)/(2)) \\ =-(-sin((\pi)/(2))) \\ =sin(\pi)/(2) \\ =1 \end{gathered}`

Also,

`\begin{gathered} \cos\beta=cos(-(7\pi)/(2)) \\ =\cos(7\pi)/(2) \\ =-cos(\pi)/(2) \\ =0 \end{gathered}`

And,

`\begin{gathered} \tan\beta=tan(-(7\pi)/(2)) \\ =-tan((7\pi)/(2)) \\ =-tan((\pi)/(2)) \\ =-\infty \end{gathered}`

Hence, the trigonometric ratios are,

`sin\beta=1,cos\beta=0,tan\beta=-\infty`