The numerical value of sin²5° + sin²10° + sin²15° +... sin²85° + sin²90° is equal a) 17/2 b) 19/2 c) 15/2 d) 13/2

Answer: [tex]b)~\Large\boxed{\frac{19}{2} }[/tex]
Step-by-step explanation:
sin²5° + sin²10° + sin²15° +... sin²85° + sin²90°
sin²x + cos²x = 1
sin(x) = cos (90 - x)
There are in total these terms:
5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 80, 85, 90
In total, there are 18 terms, and the first one matches with the second to the last one:
5 -- 85
10 -- 80
.
.
.
40 -- 50
There are 2 terms left over:
sin²45 and sin²90
sin²5 = cos² (90 - 5) = cos²85
sin²10 = cos² (90 - 10) = cos²80
.
.
.
sin²40 = cos² (90 - 40) = cos²50
i.e. sin²85 and cos²85
Using the concept of sin²x + cos²x = 1
sin²85 + cos²85 = 1
sin²80 + cos²80 = 1
.
.
.
sin²50 + cos²50 = 1
Total = (16/2) × 1 = 8 × 1 = 8
sin²45 = (sin45) (sin45) = (√2 / 2) (√2 / 2) = 1/2
sin²90 = (sin90) (sin 90) = (1) (1) = 1
[tex]8+\dfrac{1}{2} +1=\Large\boxed{\frac{19}{2} }[/tex]
Hope this helps!! :)
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