# How which Trig Identity is that?

Discussion in 'Homework Help' started by k31453, Jun 15, 2013.

May 7, 2013
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2. ### Georacer Moderator

Nov 25, 2009
5,151
1,266
This one pops in mind, but it requires a cosine on the right end. Maybe it's a typo?
Or it could be that the phase is irrelevant, but the equals sign shouldn't be there anyway.

3. ### Shagas Active Member

May 13, 2013
802
74
I think you would first have to expand the sin(1000pit) .
Because for example Sin(2x) expands into 2sinxcosx so I'm guessing this goes similarly

4. ### k31453 Thread Starter Member

May 7, 2013
54
0

Can you please show me working out !!
I tried but it keeps giving me in cos not in sin

5. ### amilton542 Active Member

Nov 13, 2010
494
64
Just make use of the trig' identities and in your final result observe the relation is a typo.

$cos(\alpha + \beta) \equiv cos(\alpha)cos(\beta) - sin(\alpha)sin(\beta)$

$cos(2\alpha) \equiv cos^2(\alpha) - sin^2(\alpha)$

$cos(2\alpha) \equiv 1 - 2sin^2(\alpha)$

$sin^2(\alpha) \equiv \frac{1}{2}[1 - cos(2\alpha)]$

$Asin^2(\omega t) \equiv \frac{A}{2}[1 - cos(2\omega t)$

$2000sin^2(500\pi t) \equiv \frac{2000}{2}[1 - cos((2)(500\pi t))] \equiv 1000[1 - cos(1000\pi t)]$

Georacer likes this.
6. ### DerStrom8 Well-Known Member

Feb 20, 2011
2,428
1,329
I think Geo is right--it's a typo. I'm pretty sure it should be cos, not sin.

-1000(-1+cos(1000(pi)t)) is the result, and that is the same as 1000(1-cos(1000(pi)t)).

EDIT: Amilton beat me to it

7. ### WBahn Moderator

Mar 31, 2012
18,087
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Two observations.