$$\begin{gathered} Find the value of...... \\ {\sin }^2{1}^0+{\sin }^2{3}^0+{\sin }^2{5}^0..........{\sin }^2{87}^0+{\sin }^2{89}^0. \\ Use the farmula a_{n=a+(n-1)d to solve the given qwestion and also trognometric identies } \end{gathered}$$

sin²1+sin²5 + sin²9+……….sin²89 =?

Here, we notice that (1, 89), (5,85)…are complementary angles.

so we combine 1st & the last term, 2nd & the last but one term……..

ie, (sin²1+sin²89) +(sin²5+ sin²85) +………..so on..

Since these angles are in AP

1,5,9,13,………89 we find the total no of terms

Tn= a+(n-1)d

89= 1+(n-1)4

88=4n-4

4n= 92

So n= 23

Since total no of terms= 23 is an odd number

Therefore, the middle term will be n+1/2 =(23+1)/2 = 12th

Now 12th term = 1+11x4 = 1+44=45

When we combine the terms in AP series

(Sin²1+sin²89) +( sin²5+sin²85)+……….sin²45

(Sin²1+cos²1)+(sin²5+ cos²5)+…….sin²45 ( we got this step using complementary angles relation sin²a= cos²(90-a)

Now since since sin²a+ cos²a= 1(fundamental identity)

Therefore we get

1+ 1 + 1+…11times………..+sin²45

= 11+ (1/√2 )²

= 11 + 1/2

= 23/2

= 11.5