# Trig Identities

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How does something like:

sin(A+B) become sinA cosB + coaA sinB ?

Likewise with:

cos(A-B) = cosA cosB + sinA sinB

ooh..& this one:

tan(A+B) = (tanA + tanB) / 1 + tanA tanB

I just don't understand how you can expand them...I'm guessing there's no distributive property in this situations...

Thanks.

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convert to imaginary numbers (exponential form) rearrange, and convert back.

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You could convert them to the imaginary form as mentioned, but a much easier way is to just draw a diagram. I've attached a little screenshot of my notes that I did quite a while ago now (over 2 years) so sorry if you can't read it

M1 (Method 1) uses the cos rule to find the side PQ. M2 uses Pythagoras to find the side PQ. When you combine the equations, you can get a derivation for cos(A-B). That was the hardest part, because from now on you can work the rest out with some simple algebra. cos(A+B) can be found by using cos(A-(-B)) which comes out quite simply. You also know that the cos graph is a transformation of the sin curve, so sin(A+B) = cos(90 - (A+B)) and from that you can then work out sin(A-B). Once you have these two equations, it's pretty easy to work out a formula for tan(A+B) because you know that tan(A+B) = sin(A+B)/cos(A+B).

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dave, I saw your methods on a site, & it was pretty cool b/c it was a java applet, and allowed you to fiddle around with the values; although I didn't get the gist of it.

Cool stuff; thanks guys;

Edward...I don't get your method though...lol

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basically

cos(x) = (exp(ix)+exp(-ix))/2

sin(x) = (exp(ix)-exp(-ix))/2i

replace x with (A+B)

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basically

cos(x) = (exp(ix)+exp(-ix))/2

sin(x) = (exp(ix)-exp(-ix))/2i

replace x with (A+B)

just like that

it's not particularly hard. you can take out factors and whatnot.

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basically

cos(x) = (exp(ix)+exp(-ix))/2

sin(x) = (exp(ix)-exp(-ix))/2i

replace x with (A+B)

I think I'm missing the bada bing

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sin over cos is tan

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