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```
prove that: cos2Acos2B + sin 2 (A - B) - sin 2 (A + B) = cos2(A + B)

```
10 years ago

```							sin2(A-B) - sin2(A+B) =sin(A+A)sin(-B-B) [using formula ,sin2a-sin2b=sin(a-b)sin(a+b)]
=sin(-2B)sin2A=-sin2Asin2B
Now LHS becomes
= cos2Acos2B-sin2Asin2B  [using formula ,cosacosb - sinasinb =cos(a+b)]
=cos(2A+2B)=cos2(A+B) = RHS
hence proved
```
10 years ago
```							Cos2ACos2B+Sin^2(A-B)-Sin^2(A+B)=Cos2(A+B) Solution:- 	L.H.S:-Cos2ACos2B+Sin^2(A-B)-Sin^2(A+B) 		=Cos2ACos2B+Sin(A-B+A+B)*Sin(A-B-A-B)	["a^2-b^2=(a+b)(a-b)"] 		=Cos2ACos2B+Sin2A*Sin(-2B) 		=Cos2ACos2B-Sin2ASin2B 		=Cos2(A+B)			["Cos(A+B)=CosACosB-SinASinB"]  Thank You Wish You A Happy Learning Trignometry.
```
10 years ago
```							LHS= cos2A.cos2B+sin(A-B+A+B).sin(A-B-(A+B))
=cos2A.cos2B+sin(A-B+A+B).sin(A-B-A-B)
=cos2A.cos2B+sin2A.sin(-2B)
=cos2A.cos2B-sin2A.sin2B
=cos(2A+2B)=cos2(A+B)=RHS

```
2 years ago
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