Question

How do you sketch a right triangle corresponding to `cottheta=5` and find the third side, then find the other five trigonometric functions?

Answer 1

`tan theta =1/5, (sin theta, cos theta)=+-1/sqrt26(1, 5)` and, correspondingly, `(csctheta, sectheta)=+-sqrt26(1, 1/5)`. The right angled `triangle OAB` keeps off negative sign.

Explanation:

#cot theta = 5 to the ratio (adjacent side)/(opposite side) = 5

Taking the opposite side length as 1 unit,

the third side = sqrt(1^2+5^2)=sqrt 26 units#.

Now, keep off the notion of associating `theta` with a right angled

triangle.

As `cot theta > 0`, `theta in Q_1 or Q_3`.

`tan theta = 1/cot theta=1/5`

As sine and cosine are both positive in `Q_1` and both negative in

`Q_3, sin theta=1/5sqrt(1-sin^2theta)`, giving sin^2theta=1/26, and so,

`sin theta = +-1/sqrt 26`,

Likewise, with the same sign,

`cos theta = +-5/sqrt 26`

Combining these two results,

`(sin theta, cos theta)=+-1/sqrt26 (1, 6)`

Correspondingly, the reciprocals

`(csc theta, sec theta) = +-sqrt 26(1. 1/5)`

The `triangle OAB` asked for has vertices O(0, 0), A(5, 0) and B(0,

1), with `angleOAB = cot^(-1)5` and the third side AB = `sqrt 26`. I

leave making graph to the reader