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)

 

[CellCoree]

I would approach this trigonometric challenge by first acknowledging that it is a statement to be proven, rather than a question to be answered. This means that I will need to provide evidence and logical reasoning to support the given statement.

To begin, I will use the trigonometric identity $\cos^2 x = \frac{1}{2}(1+\cos 2x)$ to rewrite the left-hand side of the equation:

$(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=(2+2\cos 18^{\circ}-3)(2+2\cos 54^{\circ}-3)$

Next, I will use the double angle formula for cosine, $\cos 2x = 2\cos^2 x - 1$, to further simplify the equation:

$(2+2\cos 18^{\circ}-3)(2+2\cos 54^{\circ}-3)=(2+2(2\cos^2 9^{\circ}-1)-3)(2+2(2\cos^2 27^{\circ}-1)-3)$

Simplifying further, I get:

$(2+2\cos 18^{\circ}-3)(2+2\cos 54^{\circ}-3)=(2+4\cos^2 9^{\circ}-5)(2+4\cos^2 27^{\circ}-5)$

Now, I will use the Pythagorean identity, $\cos^2 x + \sin^2 x = 1$, to rewrite the remaining cosine terms in terms of sine:

$(2+4(1-\sin^2 9^{\circ})-5)(2+4(1-\sin^2 27^{\circ})-5)$

Simplifying once more, I get:

$(1-4\sin^2 9^{\circ})(1-4\sin^2 27^{\circ})$

Next, I will use the double angle formula for sine, $\sin 2x = 2\sin x \cos x$, to rewrite the remaining sine terms:

$(1-4(2\sin 9^{\circ} \cos 9^{\circ})^2)(1-4(2\sin 27^{\circ} \cos