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Find the value of cos6 sin 24 cos 72

Find the value of cos6 sin 24 cos 72

Grade:11

2 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please fing answer to your question below

A = cos(6).sin(24).cos(72)
2sina.cosb = sin(a+b) + sin(a-b)
2sin24.cos6 = sin(24+6) + sin(24-6)
2sin24.cos6 = sin(30) + sin(18)
sin24.cos6 =\frac{ sin(30) + sin(18)}{2}
A = cos(6).sin(24).cos(72)
A = (\frac{ sin(30) + sin(18)}{2}).cos(72)
A = (\frac{ \frac{1}{2} + sin(18)}{2}).cos(72)
A = (\frac{ 1+2sin(18)}{4}).cos(72)
Now to calculate the value of sin(18)
sin(72) = 2sin(36).cos(36)
sin(72) = 2(2sin(18).cos(18)).(1-2sin^{2}(18))
sin(72) = sin(90-18) = cos(18)
cos(18) = 2(2sin(18).cos(18)).(1-2sin^{2}(18))
1 = 4sin(18)(1-2sin^{2}(18))
sin(18) = x
1 = 4x(1-2x^{2})
8x^{3} - 4x + 1 = 0
(2x-1)(4x^{2}+2x-1) = 0
x = ½ is not possible
4x^{2}+2x-1 = 0
x = \frac{\sqrt{5}-1}{4}
sin(18) = \frac{\sqrt{5}-1}{4}
A = (\frac{ 1+2sin(18)}{4}).cos(72)
A = (\frac{ 1+2sin(18)}{4}).sin(18)
A = (\frac{ 1+2.\frac{\sqrt{5}-1}{4}}{4}).\frac{\sqrt{5}-1}{4}
A = (\frac{ \sqrt{5}+1}{8}).\frac{\sqrt{5}-1}{4}
A = \frac{ 5-1}{8.4}
A = \frac{1}{8}
Shyamal
34 Points
6 years ago
=cos6.sin24cos72
=cos6.sin24.cos(90-18)
=1/2[2cos6.sin24}(sin18)
=1/2[ sin(6+24)-sin(6-24)]sin18
=1/2[sin30.sin18+sin18.sin18]
=1/2[1/2*(5^1/2-1)/2) + {(5^1/2-1)/2}^2]
After simplyfing we get,
=1/8
 

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