MHB How Can You Prove This Trigonometric Identity?

[anemone]

Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$

 

[Saitama]

Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$

Spoiler

Using the fact that
$$\cos(3x)=4\cos^3x-3\cos x\,\,\,\,(*)$$
LHS can be written as:
$$\frac{\cos 27^{\circ}}{\cos 9^{\circ}}(4\cos^2 27^{\circ}-3)=\frac{4\cos^3 27^{\circ}-3\cos 27^{\circ}}{\cos 9^{\circ}}$$
Again by using (*),
$$\frac{4\cos^3 27^{\circ}-3\cos 27^{\circ}}{\cos 9^{\circ}}=\frac{\cos 81^{\circ}}{\cos 9^{\circ}}=\frac{\sin 9^{\circ}}{\cos 9^{\circ}}=\tan 9^{\circ}$$

 

[kaliprasad]

Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$

Spoiler

we know

$\cos 3x = 4 cos ^3 x - 3 cos x$

so $\frac{\cos 3x}{\cos x} = 4 cos ^2 x - 3$

put $x = 9^{\circ}$ to get $(4\cos^2 9^{\circ}-3)= \cos 27^{\circ}/\cos 9^{\circ}$

put $x = 27^{\circ}$ to get $(4\cos^2 27^{\circ}-3)= \cos 81^{\circ}/\cos 27^{\circ}$

hence $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)$
=
$\cos 81^{\circ}/\cos 27^{\circ}\cos 27^{\circ}/\cos 9^{\circ}$
=$ \cos 81^{\circ}/\cos 9^{\circ}$
= $ \sin 9^{\circ}/\cos 9^{\circ}$
= $ tan 9^{\circ}$

 

[anemone]

Spoiler

Using the fact that
$$\cos(3x)=4\cos^3x-3\cos x\,\,\,\,(*)$$
LHS can be written as:
$$\frac{\cos 27^{\circ}}{\cos 9^{\circ}}(4\cos^2 27^{\circ}-3)=\frac{4\cos^3 27^{\circ}-3\cos 27^{\circ}}{\cos 9^{\circ}}$$
Again by using (*),
$$\frac{4\cos^3 27^{\circ}-3\cos 27^{\circ}}{\cos 9^{\circ}}=\frac{\cos 81^{\circ}}{\cos 9^{\circ}}=\frac{\sin 9^{\circ}}{\cos 9^{\circ}}=\tan 9^{\circ}$$

Spoiler

we know

$\cos 3x = 4 cos ^3 x - 3 cos x$

so $\frac{\cos 3x}{\cos x} = 4 cos ^2 x - 3$

put $x = 9^{\circ}$ to get $(4\cos^2 9^{\circ}-3)= \cos 27^{\circ}/\cos 9^{\circ}$

put $x = 27^{\circ}$ to get $(4\cos^2 27^{\circ}-3)= \cos 81^{\circ}/\cos 27^{\circ}$

hence $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)$
=
$\cos 81^{\circ}/\cos 27^{\circ}\cos 27^{\circ}/\cos 9^{\circ}$
=$ \cos 81^{\circ}/\cos 9^{\circ}$
= $ \sin 9^{\circ}/\cos 9^{\circ}$
= $ tan 9^{\circ}$

Hi Pranav and kaliprasad,:)

Thank you so much for participating in this not very difficult challenge trigonometric problem and well done! Yes, the trick to prove this identity quickly lies with the fact if one is familiar with the triple angle formula for cosine function.(Muscle)