Question

sum r=1 ^ 9 sin^ 2 r pi 18 =5​

Answer 1

Given :   ∑sin² rπ/18   =  5  where r is from 1 to 9

To Find : Prove that ∑sin² rπ/18   =  5  where r is from 1 to 9

Solution:

∑sin² rπ/18   =  5  r is from 1 to 9

LHS = ∑sin² rπ/18    r = 1 to 9

= sin²  π/18 + sin²  2π/18  + sin²  3π/18  + sin²  4π/18  + sin²  5π/18  + sin²  6π/18  + sin²  7π/18  + sin²  8π/18  + sin²  9π/18  

= sin²  π/18 + sin²  2π/18  + sin²  3π/18  + sin²  4π/18  + sin²  (π/2 - 4π/18)  + sin²  (π/2 - 3π/18)  + sin²  (π/2 - 2π/18)  + sin²  (π/2 - π/18)  + sin²   π/2  

Sin(π /2- x) = Cosx

Sin π/2 = 1

= sin²  π/18 + sin²  2π/18  + sin²  3π/18  + sin²  4π/18  +Cos²  ( 4π/18)  + Cos²  ( 3π/18)  + Cos²  ( 2π/18)  + Cos²  ( π/18)  + 1²

= (  sin²  π/18  + Cos²  ( π/18) ) + ( sin²  2π/18 +  Cos²  ( 2π/18) ) + ( sin²  3π/18 +  Cos²  ( 3π/18) ) + ( sin² 4π/18 +  Cos²  ( 4π/18) )  + 1

Sin²x + Cos²x  =1

= 1  + 1 + 1 + 1 + 1

= 5

= RHS

Hence proved

∑sin² rπ/18   =  5 r is from 1 to 9

Learn More:

If A + B + C = \frac{\pi}{2} and if none of A, B, C is an odd multiple of \frac{\pi}{2}, then prove that

brainly.in/question/6964451

{cos^{2} A . cos^{2} B}

brainly.in/question/6964151

Answer 2

Step-by-step explanation:

Given Sum r=1 ^ 9 sin^ 2 r pi 18 =5

• Now cos 2theta = 1 – 2sin^2 theta
•        Or cos 2theta + 2sin^2 theta = 1
•                      2 sin^2 theta = 1 – cos 2theta / 2
•                ∑ r = 1 to 9 sin^2 r π / 18
•               ∑ r = 1 to 9 ½ (1 – cos 2 (rπ / 18)
•               ½ ∑ r = 1 to 9 (1 – cos rπ / 9)
•                ½ [ ∑ r = 1 to 9  1  - ∑ r = 1 to 9 cos rπ / 9]
•             ½ [ 9 – (cos π / 9 + cos 2π / 9 + cos 3π / 9 + + + ++++++cos 9π / 9)
•             ½ [ 9 – (cos π / 9 + cos (π / 9 + π / 9) + cos (π/9 + 2 π/9) +--------------cos (π/9 + 8 π/9)
•            cos A + cos(A + B) + cos (A + 2B)+---------cos (A + (n – 1)B)
•                        sin nB/ 2 cos (A + (n -1)B/ 2) / sin B/2 (n = 9, A = π/9, B = π/9) / sin (π/9 x 2) )
•                   ½ [ 9 – sin (9 π / 9 x 2) cos (π/9 + 8 π/ 9 x 2) / sin (π/9 x 2)
•                     ½ [ 9 – sin (π/2 ) cos 5 π/9 / sin (π/18)
•      ½ [ 9 – 1.sin (π/2 - 5 π/9 ) / sin (π/18)]       (sin 90 = 1 , sin (90 – Ɵ) = cos Ɵ, sin (- Ɵ) = - sin Ɵ)
•                      ½ [9 – (sin - π/18 / sin π/18) ]
•                       ½ (9 – (- sin π/18) / sin π/18
•                     ½ (9 + 1)
•                    ½ (10)
•                       5

Reference link will be

https://brainly.in/question/3377161