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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...

 

Please reply ASAP.

 

Thanks.

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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).

 

Hope that answers the question.

cmpangles.jpg

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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)

 

and if you have da bada bing, you have the answer.

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Originally posted by Radical Edward

basically

 

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

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

 

replace x with (A+B)

 

and if you have da bada bing, you have the answer.

 

just like that ;)

 

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

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Originally posted by Radical Edward

basically

 

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

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

 

replace x with (A+B)

 

and if you have da bada bing, you have the answer.

 

I think I'm missing the bada bing

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