Question

cos 9 + sin9 theta is equal to root 2 Sin 54 theta​

Answer 1

Answer:

Cos(9°) + Sin(9°)  =  √2 Sin(54°)

Step-by-step explanation:

Cos(9°) + Sin(9°)  =  √2 Sin(54°)

Squaring both sides

=> (Cos(9°) + Sin(9°))²  =  (√2 Sin(54°))²

=> Cos²(9°) + Sin²(9°) + 2Cos(9°)Sin(9°)  = 2Sin²(54°)

Cos²x + Sin²x = 1

=> 1 + 2Cos(9°)Sin(9°)  = 2Sin²(54°)

=> 2Cos(9°)Sin(9°)  = Sin²(54°) + Sin²(54°) - 1

=> 2Cos(9°)Sin(9°)  = Sin²(54°) - (1 - Sin²(54°))

=> 2Cos(9°)Sin(9°)  = Sin²(54°) - (Cos²(54°))

=> 2Cos(9°)Sin(9°)  = - ((Cos²(54°) - Sin²(54°) )

Sin2x = 2SinxXosx   & Cos2x = Cos²x - Sin²x

=> Sin(18°) =  -Cos(108°)

=>  Sin(18°) =  -Cos(90° + 18°)

Cos(90° + x) = -Sinx

=> Sin(18°) = -(-Sin(18°))

=> Sin(18°) = Sin(18°)

=> LHS = RHS

QED