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I know the common one draw a line ox on x axis (straight line)mark a point y above x and zoin oy let angle xoy be a , similarlly draw a point above y mark it z join oz and let angle yoz be b.
Next join a perpendicular to ox line mark it L.
Then join zy in such a way that angle oyz is 90 degree (and a perpendicular to ox extended joining y then mark poin m thwm my=rl
In triangle zoL sin (a+b)=zL/zo
If we have a point R perpendicular to zL such that angle zRy=90degree
Then zl/zo=zr/zo +rL/zo
= zr/zo +my/zo
=zr/zo × zy/zy +my/zo × oy/oy
=zr/zy × zy/zo +my/oy × oy/zo
zr/zy would be cos a as angle rzy
Cos a×sin b+sin a as×cos b
If unable to understand then search for a diagram
Easy Math Editor
This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.
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Using Euler's formula, we can represent Sin and Cos in terms of the complex exponential:
eiθ =cosθ+isinθ
Since Sin is the imaginary part:
sinθ=Im(eiθ)
Applying to the question:
sin(A+B)=Im(ei(A+B))
Expanding the exponential:
=Im(eiA×eiB)
Rewriting in terms of Sin and Cos:
=Im[cos(A)+isin(A))×(cos(B)+isin(B))]
Expanding the product:
=Im[cos(A)cos(B)+icos(A)sin(B)+isin(A)cos(B)+i2sin(A)sin(B)]
Collecting into the real and imaginary parts:
=Im[(cos(A)cos(B)−sin(A)sin(B))+i(cos(A)sin(B)+sin(A)cos(B))
Therefore:
sin(A+B)=cos(A)sin(B)+sin(A)cos(B)
as required!
(This method also makes it really simple to find cos(A+B), as we can see from Euler's formula:
cosθ=Re(eiθ)
Therefore: cos(A+B)=cos(A)cos(B)−sin(A)sin(B)
Hope this helps!
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Thanks Ryan!!But I guess this is quite above my wavelength.Can you please provide me with an easier one??Thanks for your time anyway!!
I know the common one draw a line ox on x axis (straight line)mark a point y above x and zoin oy let angle xoy be a , similarlly draw a point above y mark it z join oz and let angle yoz be b.
Next join a perpendicular to ox line mark it L. Then join zy in such a way that angle oyz is 90 degree (and a perpendicular to ox extended joining y then mark poin m thwm my=rl In triangle zoL sin (a+b)=zL/zo If we have a point R perpendicular to zL such that angle zRy=90degree Then zl/zo=zr/zo +rL/zo = zr/zo +my/zo =zr/zo × zy/zy +my/zo × oy/oy =zr/zy × zy/zo +my/oy × oy/zo zr/zy would be cos a as angle rzy Cos a×sin b+sin a as×cos b If unable to understand then search for a diagram
This one is quite good for a simple geometrical understanding: http://demonstrations.wolfram.com/CosineAndSineOfTheSumOfTwoAngles/
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Thanks Ben!!