# Inverse trigonometric function

• October 14th 2008, 02:02 AM
maybeline9216
Inverse trigonometric function
how to do these types of qns??any tips??
• October 14th 2008, 02:12 AM
Peritus
1. (i)

$
\quad \sin \left( {2\theta } \right) = 2\cos \left( \theta \right)\sin \theta$

$

\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}$

$\cos \left( \theta \right) = \sqrt {1 - \sin ^2 \left( \theta \right)}$

$
\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
• October 14th 2008, 02:24 AM
maybeline9216

How do u get this? the 1.2 how to get did u use any exact values of trigo functions? If so, wat did u use??

http://www.mathhelpforum.com/math-he...ac410532-1.gif
Do u have any special way to rmb it?? cos it is unlike the half-angle formulae

thks for the link but i am so "overwhelmed" by it...
• October 14th 2008, 07:44 AM
NyRychVantel
Quote:

Originally Posted by maybeline9216

How do u get this? the 1.2 how to get did u use any exact values of trigo functions? If so, wat did u use??

http://www.mathhelpforum.com/math-he...ac410532-1.gif
Do u have any special way to rmb it?? cos it is unlike the half-angle formulae

thks for the link but i am so "overwhelmed" by it...

Do you know that $\sin \left( {\sin ^{ - 1} \theta } \right) = \theta$ ?
So,

$\begin{array}{l}
2\sin \left( {\sin ^{ - 1} 0.6} \right)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
= 2(0.6)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
= 1.2\cos \left( {\sin ^{ - 1} 0.6} \right) \\
\end{array}$

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.
$\cos \theta = \sqrt {1 - \sin ^2 \theta }$ is derived from $\cos ^2 \theta + \sin ^2 \theta = 1$

$\begin{array}{l}
\cos ^2 \theta = 1 - \sin ^2 \theta \\
\cos \theta = \sqrt {1 - \sin ^2 \theta} \\
\end{array}$

Similarly, for half-angle formula, consider $\sin \frac{\theta }{2}$

$\begin{array}{l}
\cos 2\theta = 1 - 2\sin ^2 \theta \\
\sin ^2 \theta = \frac{{1 - \cos 2\theta}}{2} \\
\sin \theta = \pm \sqrt {\frac{{1 - \cos 2\theta}}{2}} \\
\sin \frac{\theta}{2} = \pm \sqrt {\frac{{1 - \cos \theta}}{2}} \\
\end{array}$
• October 15th 2008, 05:08 AM
maybeline9216
Quote:

Originally Posted by NyRychVantel
Do you know that $\sin \left( {\sin ^{ - 1} \theta } \right) = \theta$ ?
So,

$\begin{array}{l}
2\sin \left( {\sin ^{ - 1} 0.6} \right)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
= 2(0.6)\cos \left( {\sin ^{ - 1} 0.6} \right) \\
= 1.2\cos \left( {\sin ^{ - 1} 0.6} \right) \\
\end{array}$

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.
$\cos \theta = \sqrt {1 - \sin ^2 \theta }$ is derived from $\cos ^2 \theta + \sin ^2 \theta = 1$

$\begin{array}{l}
\cos ^2 \theta = 1 - \sin ^2 \theta \\
\cos \theta = \sqrt {1 - \sin ^2 \theta} \\
\end{array}$

Similarly, for half-angle formula, consider $\sin \frac{\theta }{2}$

$\begin{array}{l}
\cos 2\theta = 1 - 2\sin ^2 \theta \\
\sin ^2 \theta = \frac{{1 - \cos 2\theta}}{2} \\
\sin \theta = \pm \sqrt {\frac{{1 - \cos 2\theta}}{2}} \\
\sin \frac{\theta}{2} = \pm \sqrt {\frac{{1 - \cos \theta}}{2}} \\
\end{array}$

Thanks so much!!This is really useful!! =)