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Cheers. Any help is appreciated.

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- January 30th 2009, 01:13 PMKaitosanHow do you derive the following trig identity?
http://www.mathhelpforum.com/math-he...5c42f815-1.gif

Cheers. Any help is appreciated. - January 30th 2009, 02:34 PMSoroban
Hello, Kaitosan!

Quote:

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- January 30th 2009, 08:39 PMKaitosan
Actually... I'd rather you try with the left side (I have specific reasons). Can you do that....?

This is as far as I can go -

(sinx)^2 + (cosx)^2 = 1

(sinx + cosx)^2 = 1 - sin2x

sinx + cosx = sqrt(1 - sin2x)

.................................................. ............ - January 31st 2009, 12:39 AMrunning-gag
Hi

This is not the good way to proceed.

and can be seen as the cos and the sin of an angle because the sum of their square is equal to 1

- January 31st 2009, 01:02 AMred_dog

- January 31st 2009, 01:10 AMRincewind
- January 31st 2009, 11:07 AMKaitosan
- January 31st 2009, 11:48 AMred_dog
In the second step I used the identity

- January 31st 2009, 12:22 PMKaitosan
Interesting! Could you please explain how that identity comes into being because I've never seen it? Right now I'm in calculus and I just want to make sure I know all the identities.