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(1 + Cos π/10) (1 + Cos 3π/10) (1 + Cos 7π/10) (1 + Cos 9π/10) = 1/16

(1 + Cos π/10) (1 + Cos 3π/10) (1 + Cos 7π/10) (1 + Cos 9π/10) = 1/16

Grade:10

1 Answers

Arun
25750 Points
5 years ago
The first thing to do is to realize that cos(π/10) = -cos(9π/10), 
and that cos(3π/10) = -cos(7π/10). 
In general, cos(theta) = -cos(π - theta), which is why the above are true. 

So our equation can be changed to: 
(1 + cos(π/10))(1 - cos(π/10))(1 + cos(3π/10))(1 - cos(3π/10)) 
which is: 
(1 - cos^2(π/10))(1 - cos^2(3π/10)) 
which is: 
sin^2(π/10)sin^2(3π/10) 
(1 - cos^2(theta) = sin^2(theta) for any theta... comes from the fact that cos^2(theta) + sin^2(theta) = 1) 

setting this equal to 1/16, and taking the square root of both sides: 
sin(π/10)sin(3π/10) = 1/4. 
We could also include sin(π/10)sin(3π/10) = -1/4, but this is meaningless, because both arguments (π/10 and 3π/10) are in the 1st quadrant, so sine values will be positive. 

Finding the exact values of sin(π/10) and sin(3π/10) are a little tricky... we need the multiple angle formula. If you want help with this, try googling sin(π/5) and sin(π/10) and clicking on the mathworld.wolfram.com links. They do a pretty good job of showing the proofs. 

Anyways, the exact values turn out to be: 
sin(π/10) = 1/4 (sqrt(5) - 1) 
sin(3π/10) = 1/4 (sqrt(5) + 1) 

multiplying them together gives us: 
1/4 * 1/4 * (sqrt(5) - 1)(sqrt(5) + 1) 
1/16 * (5 - 1) = 1/16 * 4 = 1/4. 

Hope this helps.

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