prove of sin(a+b)= sin(a).cos(b) + sin(b).cos(a)

[Highlander-SP]

vector prove sin(a-b)

What is the main way to prove that sin(a+b)= sin(a).cos(b) + sin(b).cos(a) ?
ps. without euler

 

[eltonjohn]

you can do it geometrically .By using two right triangles one of angle A and the other of angle B side by side and computing the sin(A +B ) od new TRIANGLE in terms of sine and cosines of A an B angles
You can also find it with a VECTOR approach and a ROTATION and find the MATRIX of such rotation it comes from the fact that ROTATIONS are linear mappings .so they verify :
R(A+ B) = R(A) + R(B) ..This is linear algebra is very simple to demostrate

 

[manitooo]

you can use euler theory I think... with the exp notation...

edit : sorry I didn't saw the euler PS...

 

[nephi]

you can draw a circle and the proof appears after some purely geometric combinations.

 

[wh_hsn]

check this page:
**broken link removed**

it has an applet to demonstrate the proof.

sin(A+B)=sin A cos B + cos A sin B
sin(A-B)=sin A cos B - cos A sin B
cos(A+B)=cos A cos B - sin A sin B
cos(A-B)=cos A cos B + sin A sin B

 

[phongphanp]

Anybody can tell me an aplication of above relation?

 

[hamidr_karami]

hi

i suggest you

refer to lithold's book


bye

 

[usernam]

Anybody can tell me an aplication of above relation?

In any number of places. Suppose u wanted to find sin75 using sin45 and sin30.

 

[eogotenks]

look in this page
https://mathforum.org/dr.math/faq/formulas/faq.triangle.html

 

[Lord Loh.]

What is the main way to prove that sin(a+b)= sin(a).cos(b) + sin(b).cos(a) ?
ps. without euler

:idea: You could use the reverse engineering method.

Start from the RHS and reach teh LSH.

There exist a formula :

2 sinA.cosB =sin(A+B)+sin(A-B)

multiply and divide the RHS by 2.

so you have

½.2.sin(a).cos(b)+½.2.sin(b).cos(a)
½{sin(a+b)+sin(a-b)}+½{sin(b+a)+sin(b-a)}
½{sin(a+b)+sin(a-b)+sin(a+b)-sin(a-b)}
½{sin(a+b)+sin(a+b)}
sin(a+b)

write these in the reverse order and you have the proof. !