Question

sin square5° + sin square6° + sin square7° +.........+ sin square 85°
(a) 10.5
(b) 40.5
(c) 1
(d) 38​

Answer 1

Solution :-

Here , we need to find the value of sin²5° + sin²6° + sin²7° + …… + sin²85°

Let's take AP :- 5° , 6° , …… 85°

Firstly finding total no. of terms using ,

• tn = a + ( n - 1 ) d

Here ,

• tn :- last term = 85°
• a :- first term = 5°
• n :- no. of terms = ?
• d :- common difference = t2 - t1 = 6° - 5° = 1°

→ 85° = 5 + ( n - 1 ) ( 1 )

→ 85° - 5° = n - 1

→ 80° + 1 ° = n

→ No. of terms = 81

Here , sin²85° can be written as sin²(90° - 5°) . So , as we know that , sin²(90° - A) = cos²A → sin²(90° - 5°) = cos²5°

• AP :-

⇒ sin²5° + sin²6° + …. + cos²6° + cos²5°

Here the middle term is ,

⇒ sin²5° + sin²6° + ….. cos²45° ….. + cos²6° + cos²5°

⇒ sin²5° + cos²5° + sin²6° + cos²6° …… cos²45° + sin²84° + cos²84° + sin²84° + sin²84°

So , as we know that sin²A + cos²A = 1 similarly ,

⇒ 1 + 1 + 1 + …… cos²45° …… + 1 + 1 + 1

There are 81 terms so by removing the term cos45° = 1/2 there will be 40 pairs of sin²A + cos²A . So ,

⇒ 40 ( 1 ) + cos²45°

⇒ 40 + ( 1/√2 )² [ °.° cos45° = 1/√2 ]

⇒ 40 + ½

⇒ 40.5

Hence , sin²5° + sin²6° + sin²7° + …… + sin²85° = 40.5 . So , option ( b ) is your answer.