trig identities

**JCM133** · August 9th 2010, 01:53 PM

I’m stuck on the following problem (see Attachment) any help would be greatly appreciated

trig identities 2.rtf

I’m not sure how it goes from here to the next line
Thanks
Kind regards

**skeeter** · August 9th 2010, 02:31 PM

you need to review your post ...

**JCM133** · August 9th 2010, 03:00 PM

Hi All
Im having a problem with the problem in the attached file

trig identities 2.rtf

any help would be greatly appreciated
Thanks Kind Regards

John

**eumyang** · August 9th 2010, 04:19 PM

$2 \left( \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cos^2 \frac{\phi}{2} + \sin \frac{\phi}{2} \cos \frac{\phi}{2} \cos^2 \frac{\theta}{2} + \sin \frac{\phi}{2} \cos \frac{\phi}{2} \sin^2 \frac{\theta}{2} + \sin \frac{\theta}{2} \cos \frac{\theta}{2} \sin^2 \frac{\phi}{2}\right)$
(I rearranged the factors in the last term to make it consistent.)

Distribute the 2:
$2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cos^2 \frac{\phi}{2} + 2 \sin \frac{\phi}{2} \cos \frac{\phi}{2} \cos^2 \frac{\theta}{2} + 2 \sin \frac{\phi}{2} \cos \frac{\phi}{2} \sin^2 \frac{\theta}{2} + 2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \sin^2 \frac{\phi}{2}$

Note that each term begins with $2\sin \frac{\theta}{2} \cos \frac{\theta}{2}$ or $2\sin \frac{\phi}{2} \cos \frac{\phi}{2}$ . Recall the sine of a double angle identity: $\sin 2u = 2\sin u \cos u$ So
$2\sin \frac{\theta}{2} \cos \frac{\theta}{2} = \sin \left( 2\cdot \frac{\theta}{2} \right) = \sin \theta$ , and the same for angle phi.

Replace:
$\sin \theta \cos^2 \frac{\phi}{2} + \sin \phi \cos^2 \frac{\theta}{2} + \sin \phi \sin^2 \frac{\theta}{2} + \sin \theta \sin^2 \frac{\phi}{2}$

Rearrange:
$\sin \theta \cos^2 \frac{\phi}{2} + \sin \theta \sin^2 \frac{\phi}{2} + \sin \phi \cos^2 \frac{\theta}{2} + \sin \phi \sin^2 \frac{\theta}{2}$

Factor:
$\sin \theta \left( \cos^2 \frac{\phi}{2} + \sin^2 \frac{\phi}{2} \right) + \sin \phi \left( \cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2} \right)$

Use the Pythagorean identity:
$\sin \theta (1) + \sin \phi (1)$

$\sin \theta + \sin \phi$

**JCM133** · August 9th 2010, 11:14 PM

Thanks EUMYANG that has helped a bundle

kind regards
John

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