how to do these types of qns??any tips??
1. (i)
$\displaystyle
\quad \sin \left( {2\theta } \right) = 2\cos \left( \theta \right)\sin \theta $
$\displaystyle
\begin{gathered}
\sin \left( {2\sin ^{ - 1} (0.6)} \right) = 2\sin \left( {\sin ^{ - 1} \left( {0.6} \right)} \right)\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) = \hfill \\
= 1.2\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) \hfill \\
\end{gathered} $
$\displaystyle \cos \left( \theta \right) = \sqrt {1 - \sin ^2 \left( \theta \right)} $
$\displaystyle
\Rightarrow 1.2\cos \left( {\sin ^{ - 1} \left( {0.6} \right)} \right) = 1.2\sqrt {1 - \sin ^2 \left( {\sin ^{ - 1} \left( {0.6} \right)} \right)} = 1.2\sqrt {1 - 0.6^2 } = 0.96$
all the other problems can be solved similarly using various trigonometric identities. So you better familiarize yourself with them. Here's a good place to start:
List of trigonometric identities - Wikipedia, the free encyclopedia
Do you know that $\displaystyle \sin \left( {\sin ^{ - 1} \theta } \right) = \theta $ ?
So,
$\displaystyle \begin{array}{l}
2\sin \left( {\sin ^{ - 1} 0.6} \right)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
= 2(0.6)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
= 1.2\cos \left( {\sin ^{ - 1} 0.6} \right) \\
\end{array}$
There's no need to remember complicated trigo identities, just remember the original identities(which is given on your formula table during your O Level exam) and manipulate it.
$\displaystyle \cos \theta = \sqrt {1 - \sin ^2 \theta } $ is derived from $\displaystyle \cos ^2 \theta + \sin ^2 \theta = 1$
$\displaystyle \begin{array}{l}
\cos ^2 \theta = 1 - \sin ^2 \theta \\
\cos \theta = \sqrt {1 - \sin ^2 \theta} \\
\end{array}$
Similarly, for half-angle formula, consider $\displaystyle \sin \frac{\theta }{2}$
$\displaystyle \begin{array}{l}
\cos 2\theta = 1 - 2\sin ^2 \theta \\
\sin ^2 \theta = \frac{{1 - \cos 2\theta}}{2} \\
\sin \theta = \pm \sqrt {\frac{{1 - \cos 2\theta}}{2}} \\
\sin \frac{\theta}{2} = \pm \sqrt {\frac{{1 - \cos \theta}}{2}} \\
\end{array}$