The trigonometric identity to prove is $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$. The key to proving this identity lies in utilizing the triple angle formula for the cosine function. Participants in the discussion, including Pranav and kaliprasad, emphasized the importance of this formula in simplifying the proof. Mastery of the triple angle formula is essential for efficiently tackling similar trigonometric challenges.
PREREQUISITESStudents, mathematicians, and educators interested in enhancing their understanding of trigonometric identities and proofs will benefit from this discussion.
anemone said:Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$
anemone said:Prove that $(4\cos^2 9^{\circ}-3)(4\cos^2 27^{\circ}-3)=\tan 9^{\circ}$
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}$