# Taking 1/2 of arcsin/arccos/arctan

• October 27th 2010, 06:21 PM
kjv6675
Taking 1/2 of arcsin/arccos/arctan
Could I get some help with the following:

sin{1/2acrsin(-7/25)}
cos(1/2arctan 8/15)
tan(1/2acrcos 3/5)

I know how to do them if the 1/2 isn't in there but I don't know how to get the answers for them. The answers btw are:

1. -1/10 * 2^(1/2)
2.
4/17 * 17^(1/2)
3.
1/2

• October 27th 2010, 08:06 PM
TheEmptySet
Quote:

Originally Posted by kjv6675
Could I get some help with the following:

sin{1/2acrsin(-7/25)}
cos(1/2arctan 8/15)
tan(1/2acrcos 3/5)

I know how to do them if the 1/2 isn't in there but I don't know how to get the answers for them. The answers btw are:

1. -1/10 * 2^(1/2)
2.
4/17 * 17^(1/2)
3.
1/2

List of trigonometric identities - Wikipedia, the free encyclopedia

This will be helpful here is the first

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

This gives

$\displaystyle \sin\left( \frac{1}{2}\sin^{-1}(\frac{-7}{25})\right)=\pm \sqrt{\frac{1-\cos(\sin^{-1}(\frac{-7}{25}))}{2}}$

$-\sqrt{\frac{1-\frac{24}{25}}{2}}=-\sqrt{\frac{1}{50}}=-\frac{\sqrt{2}}{10}$
• October 27th 2010, 08:24 PM
sa-ri-ga-ma
$Let \frac{1}{2}arcsin(\frac{-7}{25}) = \theta$

$arcsin(\frac{-7}{25}) = 2\theta$

$(\frac{-7}{25}) = \sin{2 \theta}$

$(\frac{-7}{25}) = \frac{2\tan{\theta}}{1 + \tan^2{\theta}}$

Simplify and solve the quadratic to find tanθ. from that find sinθ.

Similarly try the other problems.
• October 27th 2010, 08:25 PM
sa-ri-ga-ma
$Let \frac{1}{2}arcsin(\frac{-7}{25}) = \theta$

$arcsin(\frac{-7}{25}) = 2\theta$

$(\frac{-7}{25}) = \sin{2 \theta}$

$(\frac{-7}{25}) = \frac{2\tan{\theta}}{1 + \tan^2{\theta}}$

Simplify and solve the quadratic to find tanθ. from that find sinθ.

Similarly try the other problems.
• October 28th 2010, 04:13 AM
kjv6675
Half angle formula...thanks! I don't know why I didn't think of that. (Doh)