few sums...!!!!!!

2) sin⁴Ï€/6 + 1 + sin⁴Ï€/6 + sin⁴Ï€/6 = 1/8 + 1 + 1/8 + 1/8 = 11/8 why 3/2?

6) cos 2A = cos[(A + B) + (A - B)] = cos(A + B).cos(A - B) - sin(A + B).sin(A - B)

Now, cos(A + B) = 4/5 so, sin(A + B) = 3/5
and, sin(A - B) = 5/13 so, cos(A - B) = 12/13

So, cos2A = 4/5.12/13 - 3/5.5/13 = (48 - 15)/65 = 13/65 = 1/5

so, tan2A = 2√6 so, 2tan2A = 4√6 Ans.

IN Q 2, all the denominators will be 8 like sin(pi/8) etc..

Q.7 The max. value is 5.5
The min. value is 1

Q.2) take LHS
sin⁴(Ï€/6)+sin⁴(3Ï€/6)+sin⁴(5Ï€/6)+sin⁴(7Ï€/6)
=sin⁴(Ï€/2-Ï€/3)+sin⁴Ï€/2+sin4⁴(Ï€/2+Ï€/3)+sin⁴(Ï€/2+2Ï€/3)
=cos⁴Ï€/3+sin⁴Ï€/2+sin⁴Ï€/3+sin⁴2Ï€/3
=1+1+9/16≠3/2

even if i have made any calculation mistakes to get 3/2 u need π/8 NOT π/6