Question

Prove that cos theta.cos theta/2-cos 3 theta.cos9 theta/2= sin 7 theta.sin 8 theta​

Answer 1

Step-by-step explanation:

Text Solution

Solution :

LHS = cosθcosθ2−cos3θcos9θ2

=12[2cosθ⋅cosθ2−2cos3θ⋅cos9θ2]

=12[cos(θ+θ2)+cos(θ−θ2)−cos(3θ+9θ2)−cos(3θ−9θ2)]

12(cos3θ2+cosθ2−cos15θ2−cos3θ2

=12[cosθ2−cos15θ2]

=−12[2sin(θ+15θ2)⋅sin(θ−15θ2)] [∵cosx−cosy=−2sinx+y2⋅sinx−y2]

=+(sin8θ⋅sin7θ)=RHS

∴ LHS=RHS Hence proved.

Answer 2

Answer:

➩x×0.74=5.92

➩x=0.745.92

➩x=8

Answer:otherno.=8