Question

How do you evaluate `sec( - π / 18) times cos( (37π)/18)`?

Answer 1

1

Explanation:

`P = sec (-pi/18).cos ((37pi)/18) = (1/cos (-pi/18))cos ((37pi)/18)`
`cos ((37pi)/18) = cos (pi/18 + 2pi) = cos (pi/18)`
Finally, we get:
`P = cos (pi/18)/(cos ((-pi)/18))`
Since, on the unit circle, `cos (pi/18) = cos (-pi/18)`, therefor
P = 1