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how to verify an identity that contains the operation [+-] (stacked over each other)??this qustion for example

Guest Apr 7, 2015

Best Answer 

 #3
avatar+89791 
+5

Using a sum-to-product identity, we have

sin x + sin y = 2sin[(x + y)/2] cos[(x - y)/2]

And

cos x + cos y =  2cos[(x + y)/2]cos[(x - y)/2]

So

[sin x + sin y] / [cos x + cos y]  = 

[2sin[(x + y)/2] cos[(x - y)/2] ] /   [2cos[(x + y)/2]cos[(x - y)/2] ] =

sin [(x + y)/2] / cos[ (x + y)/2 ]    and since (x + y)/2  is just an angle, θ, then

sinθ / cosθ  = tanθ   =  tan[(x + y}/2]

 

The other part can be proved similarly by noting that

sinx - sin y  = 2cos[(x + y)/2] sin[(x - y)/ 2]

 

  

CPhill  Apr 9, 2015
 #2
avatar+890 
+5

It depends on what earlier identities that you are allowed to assume.

 

From basics,

 

sin(A+B) = sinA.cosB + cos A.sinB, and sin(A-B) = sinA.cosB - cosA.sinB

(or do we have to derive these ?)

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

Now let A+B = C and A-B = D, then A=(C+D)/2 and B=(C-D)/2,

so sinC + sinD = 2sin((C+D)/2).cos((C-D)/2).   (Are we allowed to assume this ?)

 

Now go through a similar routine starting with cos(A+B) and cos(A-B), to arrive at

cosC + cosD = 2cos((C+D)/2).cos((C-D)/2).

 

Apply these two identities to the lhs of your problem identity and the rhs (for the positive sign) drops out.

To deal with the negative sign in the middle, repeat the calculation starting with sin(A+B) - sin(A-B).

Bertie  Apr 8, 2015
 #3
avatar+89791 
+5
Best Answer

Using a sum-to-product identity, we have

sin x + sin y = 2sin[(x + y)/2] cos[(x - y)/2]

And

cos x + cos y =  2cos[(x + y)/2]cos[(x - y)/2]

So

[sin x + sin y] / [cos x + cos y]  = 

[2sin[(x + y)/2] cos[(x - y)/2] ] /   [2cos[(x + y)/2]cos[(x - y)/2] ] =

sin [(x + y)/2] / cos[ (x + y)/2 ]    and since (x + y)/2  is just an angle, θ, then

sinθ / cosθ  = tanθ   =  tan[(x + y}/2]

 

The other part can be proved similarly by noting that

sinx - sin y  = 2cos[(x + y)/2] sin[(x - y)/ 2]

 

  

CPhill  Apr 9, 2015
 #4
avatar+93627 
0

Thanks Chris and Bertie  

Melody  Apr 9, 2015

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