Question

From where this formula came from: Derive

$tan(A+B) /2=√1-cos(A-B) /√1+cos(A-B)$​

Answer 1

Answer:

cos(45−30)=(cos45)(cos30)+(sin45)(sin30)=

(2–√/2)(3–√/2)+(2–√/2)(1/2)=

(1+3–√)/(22–√)=

2–√(1+3–√)/(2(2–√)2)=

(2–√+6–√)/4

Second solution:

Let cos15=x. Then, we have:

sin30=2(sin15)(cos15)=>1/2=2(1−x2−−−−−√)(x)=>

1/4=x(1−x2−−−−−√)=>(x2)(1−x2)=1/16=>

x2−x4=1/16=>x4−x2+1/16=0…(1)

Now, we set x2=y and we take:

y2−y+1/16=0…(2)

The solutions of (2) are:

y=(1−3/4−−−√)/2 and y=(1+3/4−−−√)/2