Half-Angle Formulas

How the real exam tests it

Questions from real CSCA papers that test this concept. Work each one out before you open the answer.

1. CSCA Math December 2025 Past Paper
   Given $\cos\alpha=-\frac{1}{2}$ and $\alpha\in(\frac{\pi}{2},\pi)$, find $\sin\frac{\alpha}{2}$.
   • A．$-\frac{\sqrt{3}}{2}$
   • B．$\frac{\sqrt{3}}{2}$
   • C．$\frac{1}{2}$
   • D．$-\frac{1}{2}$
   Show answer & solution

   Answer: B

   Answer: B Since $\alpha/2\in(\pi/4,\pi/2)$, $\sin\frac{\alpha}{2}=\sqrt{\frac{1-\cos\alpha}{2}}=\frac{\sqrt{3}}{2}$. Choose B.
   See it in the full CSCA Math December 2025 Past Paper
2. CSCA Math April 2026 Past Paper
   If $\cos \alpha = - \frac { 7 } { 25 }$ and $\frac { \pi } { 2 } < \alpha < \pi$, then $\cos \frac { \alpha } { 2 } =$
   • A．$- \frac { 4 } { 5 }$
   • B．$\frac { 4 } { 5 }$
   • C．$\frac { 3 } { 5 }$
   • D．$- \frac { 3 } { 5 }$
   Show answer & solution

   Answer: C

   $\frac { \alpha } { 2 } \in \left( \frac { \pi } { 4 } , \frac { \pi } { 2 } \right)$, so $\cos \frac { \alpha } { 2 } > 0$. Half-angle: $\cos \frac { \alpha } { 2 } = \sqrt { \frac { 1 + \cos \alpha } { 2 } } = \sqrt { \frac { 18 / 25 } { 2 } } = \sqrt { \frac { 9 } { 25 } } = \frac { 3 } { 5 }$.
   Take the full CSCA Math April 2026 Past Paper — timed & graded
3. CSCA Math June 2026 Past Paper
   If $\alpha \in \left[ \frac{3\pi}{2}, 2\pi \right]$, and $\cos \alpha = \frac{3}{5}$, then $\sin \frac{\alpha}{2} =$
   • A．$-\frac{\sqrt{5}}{5}$
   • B．$\frac{\sqrt{5}}{5}$
   • C．$-\frac{2\sqrt{5}}{5}$
   • D．$\frac{2\sqrt{5}}{5}$
   Show answer & solution

   Answer: B

   $\alpha \in \left[ \frac{3\pi}{2}, 2\pi \right] \Rightarrow \frac{\alpha}{2} \in \left[ \frac{3\pi}{4}, \pi \right]$, so $\sin \frac{\alpha}{2} \geq 0$. By the half-angle formula, $\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}} = \sqrt{\frac{1 - \frac{3}{5}}{2}} = \sqrt{\frac{1}{5}} = \frac{\sqrt{5}}{5}$. Choose (B).
   Take the full CSCA Math June 2026 Past Paper — timed & graded