Question

the value of sin 1 .sin 3.sin 5 ........sin89​

Answer 1

Answer:

89/2

Look it can be done like this!.

sin(1)*sin(3)…..sin(89)=sin(1)*sin(2)*sin(3)*sin(4)..*sin(89)/sin(2)*sin(4)*sin(6)…..sin(88).

Now sin(1)*sin(2)*sin(3)..sin(89)=(sin1*sin(89)*(sin(2)*sin(88)…sin(44)*sin(46)*sin(45)= (sin2*sin(4)*…sin(88) *sin(45)) *2^(-44)

now sin(1)*(sin(3)*sin(5)…sin(89)= sin(1)*sin(2)*sin(3)*sin(4)..*sin(89)/sin(2)*sin(4)*sin(6)…..sin(88).

Since sin(1)*sin(2)*sin(3)..sin(89)=(sin1*sin(89)*(sin(2)*sin(88)…sin(44)*sin(46)*sin(45)= 2^(-44)*sin2*sin(4)*…sin(88)*sin(45)

therefore sin(1)*sin(3)…..sin(89)=2^(-44)*sin(45)=2^(-89/2). therefore n=89/2

Step-by-step explanation:

Hope it helps you ✌️