The exact value of a trig expression

Dean54321

New member
Joined
Apr 6, 2021
Messages
12
Find the exact value of sin(π/9)sin(2π/9)sin(4π/9)

when attempting to solve the equation, I tried to use product to sums, 2sin(A)sin(B) = cos(A-B) - cos(A+B) and 2sin(A)cos(B) = sin(A+B) + sin(A-B)
so, '
= 1/2sin(4π/9) x (2sin(π/9)sin(2π/9))
= 1/2sin(4π/9) x [cos(π/9-2π/9) - cos(π/9+2π/9)]
= 1/2sin(4π/9) x [cos(-π/9) -cos (3π/9)]
= 1/2sin(4π/9) x [cos(π/9) -cos (π/3)]
= 1/2sin(4π/9)cos(π/9) -1/2sin(4π/9)cos (π/3)
= 1/4(2sin(4π/9)cos(π/9))-1/4(2sin(4π/9)cos (π/3))
= 1/4[ sin(4π/9 + π/9) + sin(4π/9 + π/9) ] -1/4[ sin(4π/9+π/3) + sin(4π/9-π/3) ]
= 1/4sin(5π/9) + 1/4sin(3π/90 -1/4sin(7π/9) -1/4sin(π/9)

From here I see no simplification. However, the answer is still right, just not in exact form. Maybe I should try without making cos(-π/9) = cos(π/9), in any case I'm stuck and I really appreciate anyone who tries to help. Thank you.

[Answer = root(3)/8]
 

Poliagapitos

New member
Joined
Jul 8, 2021
Messages
12
Hint: Use the following formulae: [imath]\sin 2\alpha=2\sin \alpha \cos \alpha[/imath] [imath]\quad ; \quad[/imath] [imath]\cos 2 \alpha = \cos^2 \alpha -\sin^2 \alpha[/imath]
 
Top