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Obtaining the identity

  • Thread starter Dainy
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  • #1
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yes well I got the last problem.....but I still wanna have an idea were the identity siny+sinx=2sin(x+y)/2cos(x+y)/2 came from please someone sort of explain plz! thanx
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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Dainy said:
yes well I got the last problem.....but I still wanna have an idea were the identity siny+sinx=2sin(x+y)/2cos(x+y)/2 came from please someone sort of explain plz! thanx
what you wrote is not an identity consider x=y=pi/2
sin pi/2+sin pi/2=2
2sin pi/2 cos pi/2=0
do you mean
[tex]\sin(y)+\sin(x)=2\cos(\frac{y-x}{2})\sin(\frac{y+x}{2})[/tex]
if so
start on the left by writing
y=(y+x)/2+(y-x)/2
x=(y+x)/2-(y-x)/2
then expand using
sin(a+b)=sin(a)cos(b)+cos(a)sin(b)
and
sin(a-b)=sin(a)cos(b)-cos(a)sin(b)
add like terms
 
Last edited:
  • #3
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back to the root

Dainy:

Make sure that you understand why

cos(a-b) = cos(a) cos(b) + sin(a) sin(b)

and that you can illustrate the meaning of this formula with a drawing.

Many other formulas can be derived form the previous by algebra and by other simple trigonometric rules (likes cos(-b)=cos(b), sin(-b)=-sin(b), cos(pi-b)=-cos(b), ... all rules that can be illustrated by a drawing too.).

So the picture is: a few simple principles and definitions, enough algebra, and you are on your own.
 

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