Inverse trigonometric function

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October 14th 2008, 02:02 AM
maybeline9216

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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

http://www.mathhelpforum.com/math-he...7ad2dd16-1.gif

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

http://www.mathhelpforum.com/math-he...7ad2dd16-1.gif

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!! =)