Question

Sin6+sin12+sin18+.... +sin90=

Answer 1

Answer:

sin3θ=3sinθ－4(sinθ) ^3

=-4sinθ{(sinθ) ^2－3/4}  

=-4sinθsin(θ+60°)sin(θ-60°)

cos3

cos3θ=4sin(90°-θ)sin(30°-θ)sin(30°-θ)

sin6θ=2sin3θcos3θ

=32sinθsin(30°-θ) sin(30°+θ) sin(60°-θ) sin(60°+θ)sin(90°-θ)

Step-by-step explanation: