MHB How Can You Prove This Trigonometric Identity?

AI Thread Summary
The discussion focuses on proving the trigonometric identity $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$. Participants highlight that familiarity with the triple angle formula for cosine is key to solving the problem efficiently. The problem is considered manageable, with encouragement for those engaging with it. Overall, the conversation emphasizes the importance of understanding trigonometric identities and formulas in solving such challenges. Mastery of these concepts is essential for proving complex trigonometric identities.
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Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$
 
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anemone said:
Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$

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 (*),
$$\frac{4\cos^3 27^{\circ}-3\cos 27^{\circ}}{\cos 9^{\circ}}=\frac{\cos 81^{\circ}}{\cos 9^{\circ}}=\frac{\sin 9^{\circ}}{\cos 9^{\circ}}=\tan 9^{\circ}$$
 
anemone said:
Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$

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}$
 
Pranav said:
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 said:
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)
 
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