Verify the Identity
(sin m +sin n)/ (cos m + cos n) = tan(m + n/2)
I suspect you mean
tan((m + n)/2)
on the RHS
yes i do.
Use Lemma
sin m + sin n = 2sin(m+n/2)cos(m-n/2)
cos m + cos n = 2cos(m+n/2)cos(m-n/2)
We get
(sin m +sin n)/cos m + cos n) = sin(m+n/2)/cos(m+n/2) = tan m+n/2)
To verify the given identity:
(sin m + sin n)/(cos m + cos n) = tan(m + n/2)
we'll use the trigonometric identity:
sin(a) + sin(b) = 2sin((a + b)/2)cos((a - b)/2)
and
cos(a) + cos(b) = 2cos((a + b)/2)cos((a - b)/2)
Let's break down the proof step-by-step:
Start with the left side of the given identity:
(sin m + sin n)/(cos m + cos n)
Using the identity for the sum of sines:
= 2sin((m + n)/2)cos((m - n)/2) / (cos m + cos n)
Now, let's use the identity for the sum of cosines:
= (2sin((m + n)/2)cos((m - n)/2) / 2cos((m + n)/2)cos((m - n)/2)
The 2's cancel out:
= sin((m + n)/2) / cos((m + n)/2)
Since tan(x) = sin(x) / cos(x):
= tan((m + n)/2)
Which is equal to the right side of the given identity.
Therefore, we have verified that:
(sin m + sin n)/(cos m + cos n) = tan(m + n/2)