Question

cos 36° . cos 72° . cos 108° . cos 144° = 1/16

Answer 1

Hey there!

To prove : Cos(36) * Cos(72) * Cos(108) * Cos(144) = $\bf{\frac{1}{16}}$.

Let us solve it part by part and simply as a whole to prove the right hand side.

Taking Left hand side to proceed with this trigonometric values:

For "Cos(72)" :

Using the following identity of "Cos(x)" that is, Cos(x) = Sin(90 - x).

= Sin(90 - 72)

= Sin(18)

Take sin value as fractional form,

$\bf{= sin(\frac{36}{2})}$

Now, by applying the basic principles of the half angle identities that is,

$\bf{Sin(\frac{x}{2}) = \sqrt{\frac{1 - Cos(x)}{2}}$

Here,

$\bf{\sqrt{\frac{1 - Cos(36)}{2}}$

Now, Cos(36) - Sin(18) = $\bf{\frac{1}{2}}$.

And, Cos(36) + Sin(18) = $\bf{\sqrt{\frac{5}{4}}}$.

To obtain the value of Cos(36) = $\bf{\frac{\sqrt{5} + 1}{4}}$

$\bf{= \sqrt{\frac{1 - \frac{\sqrt{5} + 1}{4}}{2}}$

$\bf{= \frac{\sqrt{2} \sqrt{3 - \sqrt{5}}{4}}$

Similarly for Cos(108) = $\bf{= - \frac{\sqrt{2} \sqrt{3 - \sqrt{5}}{4}}$      

[Hint: Use the identity of Cos(x) = Sin(90 - x) to get Sin(- 18) = -Sin(18);  For Cos(108) and continue the process]

Similarly for Cos(144) = Cos(2 * 72) = Cos^2(36) - Sin^2(36)

For Cos(36) = $\bf{\frac{\sqrt{5} + 1}{4}}$

For Sin(36) = $\bf{\frac{\sqrt{\frac{5 - \sqrt{5}}{2}}{2}}$

Therefore, Cos(144) = $\bf{- \frac{\sqrt{5} - 1}{4}}$

Similarly for Cos(36) = $\bf{\frac{\sqrt{5} + 1}{4}}$

Add all the values for sin and cos trigonometric functions:

$\bf{\frac{1 + \sqrt{5}}{4} (- \frac{1 + \sqrt{5}}{4})}$ $\bf{\frac{\sqrt{2} \sqrt{3 - \sqrt{5}}{4}}$ $\bf{(- \frac{\sqrt{2} \sqrt{3 - \sqrt{5}}{4})}$

$\huge{\bf{\boxed{= \frac{1}{16}}}}$

Hope this helps you!

Answer 2

cos 36 = $[\sqrt{5} + 1]/4$