Question

Show that the given trigonometric equation holds true
$\dfrac{{\cos 2A + 2\cos 4A + \cos 6A}}{{\cos A + 2\cos 3A + \cos 5A}} = \cos A - \sin A.\tan 3A$

Answer 1

Hint – In this question consider the L.H.S part and apply the trigonometric identity$\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$, and do similar simplification in the denominator part. This will help getting to the desired R.H.S.
Complete step-by-step answer:
Proof –
Consider L.H.S
$ \Rightarrow \dfrac{{\cos 2A + 2\cos 4A + \cos 6A}}{{\cos A + 2\cos 3A + \cos 5A}}$
Now rearrange the terms we have,
$ \Rightarrow \dfrac{{\cos 6A + \cos 2A + 2\cos 4A}}{{\cos 5A + \cos A + 2\cos 3A}}$
Now as we know that $\cos C + \cos D = 2\cos \left( {\dfrac{{C + D}}{2}} \right)\cos \left( {\dfrac{{C - D}}{2}} \right)$ so use this property we have,
\[ \Rightarrow \dfrac{{2\cos \left( {\dfrac{{6A + 2A}}{2}} \right)\cos \left( {\dfrac{{6A - 2A}}{2}} \right) + 2\cos 4A}}{{2\cos \left( {\dfrac{{5A + A}}{2}} \right)\cos \left( {\dfrac{{5A - A}}{2}} \right) + 2\cos 3A}}\]
Now simplify the above equation we have,
\[ \Rightarrow \dfrac{{2\cos 4A\cos 2A + 2\cos 4A}}{{2\cos 3A\cos 2A + 2\cos 3A}}\]
Now take 2 cos 4A from numerator and 2cos 3A from denominator we have,
\[ \Rightarrow \dfrac{{2\cos 4A\left( {\cos 2A + 1} \right)}}{{2\cos 3A\left( {\cos 2A + 1} \right)}}\]
$ \Rightarrow \dfrac{{\cos 4A}}{{\cos 3A}}$
Now above equation is also written as
 $ \Rightarrow \dfrac{{\cos \left( {3A + A} \right)}}{{\cos 3A}}$
Now as we know that $\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B$ so use this property we have,
$ \Rightarrow \dfrac{{\cos 3A\cos A - \sin 3A\sin A}}{{\cos 3A}}$
$ \Rightarrow \cos A - \sin A.\dfrac{{\sin 3A}}{{\cos 3A}}$
Now as we know that (sin/cos) = tan so use this property we have,
$ \Rightarrow \cos A - \sin A.\tan 3A$
= R.H.S
Hence proved.

Note – Tricky point while solving such trigonometric questions is simply to understand which trigonometric identity is to be used, some of them are being mentioned above. Here we have taken out the common part in the numerator separately and the same is being done to the denominator part so as to cancel out terms. It is advised to grasp trigonometric identities, as it helps to save a lot of time.