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Jan 10, 2019
0
#3
Mr_Alex_Pims wrote:
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i need an answer
Jan 20, 2019
0
#4
FC_BayernMunich wrote:
Given that - sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B,
show that cos A sin B = 1 2 [sin(A + B) − sin(A − B)] and sin2 A = 1 2 [1 − cos 2A].
xD I hate these kind of trigonometry problems
Jan 20, 2019
0
#5
Leon_Likes_Chess wrote:
FC_BayernMunich wrote:
Given that - sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B,
show that cos A sin B = 1 2 [sin(A + B) − sin(A − B)] and sin2 A = 1 2 [1 − cos 2A].
xD I hate these kind of trigonometry problems
I know, right! They are so annoying!
Jan 20, 2019
0
#7
LetsGoPokemon wrote:
Not even in middle school
I am, I just decided to speed up my math. Didn't work out very well. I can solve these things but I suck at proving or "showing" them
Jan 20, 2019
0
#8
Leon_Likes_Chess wrote:
LetsGoPokemon wrote:
Not even in middle school
I am, I just decided to speed up my math. Didn't work out very well. I can solve these things but I suck at proving or "showing" them
I am not and i decided to do the same. I self-taught myself. I got this from a 4th year univeristy challenge problems sheet online. I finished the 3rd year stuff
Given that - sin(A ± B) = sin A cos B ± cos A sin B and cos(A ± B) = cos A cos B ∓ sin A sin B,
show that cos A sin B = 1 2 [sin(A + B) − sin(A − B)] and sin2 A = 1 2 [1 − cos 2A].