Calculating the exact value of trig ratios

arccos-1/2


I understand that because cos is negative, the angle is in either Sin/Tan quadrant. Why have they used the Sin quadrant in the question and not the tan one?

arccos-1/2

https://9d7fc71c7b4ac6b796d66d68d061a8e00d3191c7.googledrive.com/host/0B1ZiqBksUHNYU1BlRFpLN3lyYXM/CH6.pdf

I want to use the method used on P68 Qe)

I understand that because cos is negative, the angle is in either Sin/Tan quadrant. Why have they used the Sin quadrant in the question and not the tan one?

you mean $\arccos(\frac{-1}{2})$ right?

'I understand that because cos is negative, the angle is in either Sin/Tan quadrant.' - Correct

'Why have they used the Sin quadrant in the question and not the tan one?' - As i am sure you are aware there are infinity many $\theta : cos(\theta)=\frac{-1}{2}$. What you are looking for is the principle value which for $cos(\theta)$ means $0\le \theta \le\pi$. This clearly means the angle lines in the 'Sin quadrant'.

Thanks!

Could you help me with Qf)

Why is the answer -pi/6? Why haven't they applied a similar method to Qe) where you do 180-30 to give you there answer? Why have they left it as -pi/6?

Again principle values - look it up