Trigonometry> Find the value of cos6 sin 24 cos 72

Find the value of cos6 sin 24 cos 72

laxman , 10 Years ago

Grade 11

2 Answers

Jitender Singh

Last Activity: 10 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
$4x^{2}+2x-1 = 0$
$x = \frac{\sqrt{5}-1}{4}$
$sin(18) = \frac{\sqrt{5}-1}{4}$
$A = (\frac{ 1+2sin(18)}{4}).cos(72)$
$A = (\frac{ 1+2sin(18)}{4}).sin(18)$
$A = (\frac{ 1+2.\frac{\sqrt{5}-1}{4}}{4}).\frac{\sqrt{5}-1}{4}$
$A = (\frac{ \sqrt{5}+1}{8}).\frac{\sqrt{5}-1}{4}$
$A = \frac{ 5-1}{8.4}$
$A = \frac{1}{8}$