Sum and Difference Identities
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Key Questions

There are 2 identities you must know to simplify your expressions and solve equations
 sum identities :
#(a+b)^2= a^2+b^2+2*a*b#  difference identities:
#(ab)^2= a^2+b^22*a*b#
Proof:
#(a+b)^2= (a+b)*(a+b)# then you expand
#(a+b)^2=a^2+b^2+a*b+b*a# and then#a*b=b*a# #(ab)^2= (ab)*(ab)# then you expand
#(ab)^2=a^2+b^2a*bb*a# Example:
#(x+3)^2= x^2+9+6x#
#(x3)^2= x^2+96x# Bonus:
#(a+b)(ab)= a^2b^2#
proof:
#(a+b)(ab)=a^2a*b+b*ab^2=a^2b^2#  sum identities :

Here is an example:
sin(75) = sin(30 + 45) = sin(30) cos(45) + cos(30) sin(45)
= (1/2) (#sqrt2# /2) + (#sqrt3# /2)(#sqrt2# /2)
= (#sqrt2# +#sqrt6# )/4cos(75) = cos(30 + 45) = cos(30) cos(45)  sin(30) sin(45)
= (#sqrt3# /2)(#sqrt2# /2)  (1/2) (#sqrt2# /2)
= (#sqrt6# #sqrt2# )/4sin(15) = sin(45  30) = sin(45) cos(30)  cos(45) sin(30)
= (#sqrt2# /2) (#sqrt3# /2)  (#sqrt2# /2)(1/2)
= (#sqrt6# #sqrt2# )/4cos(15) = cos(45  30) = cos(45) cos(30) + sin(45) sin(30)
=(#sqrt2# /2) (#sqrt3# /2) + (#sqrt2# /2) (1/2)
=(#sqrt6# +#sqrt2# )/4 
Main Sum and Differences Trigonometric Identities
#cos (a  b) = cos a*cos b + sin a*sin b#
#cos (a + b) = cos a*cos b  sin a*sin b#
#sin (a  b) = sin a*cos b  sin b*cos a#
#sin (a + b) = sin a*cos b + sin b*cos a#
#tan (a  b) = (tan a  tan b)/(1 + tan a*tan b)#
#tan (a + b) = (tan a + tan b)/(1 tan a*tan b)# Application of Sum and Differences Trigonometric Identities
Example 1: Find
#sin 2a# .#sin 2a#
#= sin (a + a)#
#= sin a*cos a + sin a*cos a#
#= 2*sin a*cos a# Example 2: Find
#cos 2a# .#cos 2a#
#= cos (a + a)#
#= cos a*cos a  sin a*sin a#
#= cos^2 a  sin^2 a# Example 3: Find
#cos ((13pi)/12)# .#cos ((13pi)/12)#
#= cos (pi/3 + (3pi)/4)#
#= cos (pi/3)*cos ((3pi)/4)  sin (pi/3)*sin ((3pi)/4)#
#= (sqrt2)/4  (sqrt6)/4#
#= [sqrt2 + sqrt6]/4# 
Here is an example of using a sum identity:
Find
#sin15^@# .If we can find (think of) two angles
#A# and#B# whose sum or whose difference is 15, and whose sine and cosine we know.#sin(AB)=sinAcosBcosAsinB# We might notice that
#7560=15#
so#sin15^@=sin(75^@60^@)[email protected]^@[email protected]^@# BUT we don't know sine and cosine of
#75^@# . So this won't get us the answer. (I included it because when solving problems we DO sometimes think of approaches that won't work. And that's OK.)#4530=15# and I do know the trig functions for#45^@# and#30^@# #sin15^@=sin(45^@30^@)[email protected]^@[email protected]^@# #=(sqrt2/2)(sqrt3/2)(sqrt2/2)(1/2)# #=(sqrt6  sqrt 2)/4# There are other way of writing the answer.
Note 1
We could use the same two angles and the identity for#cos(AB)# to find#cos 15^@# Note 2
Instead of#4530=15# we could have used#6045=15# Note 3
Now that we have#sin 15^@# we could use#60+15=75# and#sin(A+B)# to find#sin75^@# . Although if the question had been to find#sin75^@, I'd probably use # 30^@# and # 45^@#