trig: missing steps?

Printable View

Apr 27th 2013, 03:21 PM
infraRed

trig: missing steps?

Hi. One of my math books seems to indicate that

$\sqrt {2 - 2\,sin\,t\,sin\,5t - 2\,cos\,t\,cos\,5t$

is equal to

$\sqrt {2 - 2\,cos\,4t}$

The transformation is I imagine quite obvious to everyone else, but if somebody could bear with me and show the hidden steps, it would help my learning.
Apr 27th 2013, 03:54 PM
Soroban

Re: trig: missing steps?

Hello, infraRed!

We are expected to know this identity:

. . $\cos(A\pm B) \:=\:\cos A\cos B \mp\sin A\sin B$

Quote:

My math book indicates that: . $\sqrt {2 - 2\sin t\sin 5t - 2\cos t\cos5t} \;=\; \sqrt {2 - 2\cos4t}$

The left side has:

$2 - 2\sin t\sin5t - 2\cos t\cos5t$

. . $=\;2 - 2(\cos 5t\cos t + \sin5t\sin t)$

. . $=\;2 - 2\cos(5t-t)$

. . $=\;2 - 2\cos4t$

All times are GMT -8. The time now is 09:00 PM.

Copyright © 2005-2017 Math Help Forum. All rights reserved.