Question

Prove the following: \[frac{{\cos 3\theta + \cos 3\phi }}{{2\cos (\theta - \phi ) - 1}} = (\cos \theta + \cos \phi )\cos (\theta + \phi ) - (\sin \theta + \sin \phi )\sin (\theta + \phi )\]

Answer 1

Hint: Simplify the left-hand side of the equation using sum of cosine formula \[\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\] and formula for cos2x, that is, \[\cos 2x = 2{\cos ^2}x - 1\]. Then, simplify the right-hand side of the equation using the formula \[\cos A\cos B - \sin A\sin B = \cos (A + B)\] and then complete the proof.

Complete step by step answer:
Let the left-hand side of the equation be LHS.

\[LHS = \dfrac{{\cos 3\theta + \cos 3\phi }}{{2\cos (\theta - \phi ) - 1}}............(1)\]

We know that the formula for the sum of cosines is given as follows:

\[\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\]

Using this formula in equation (1), we have:

\[LHS = \dfrac{{2\cos \left( {\dfrac{{3\theta + 3\phi }}{2}} \right)\cos \left( {\dfrac{{3\theta - 3\phi }}{2}} \right)}}{{2\cos (\theta - \phi ) - 1}}\]

\[LHS = \dfrac{{2\cos \left( {3\left( {\dfrac{{\theta + \phi }}{2}} \right)} \right)\cos \left( {3\left( {\dfrac{{\theta - \phi }}{2}} \right)} \right)}}{{2\cos (\theta - \phi ) - 1}}............(2)\]

We know the formula for cos3x, given as follows:

\[\cos 3x = 4{\cos ^3}x - 3\cos x\]

Using this formula in equation (2), we get:

\[LHS = \dfrac{{2\cos \left( {3\left( {\dfrac{{\theta + \phi }}{2}} \right)} \right)\left[ {4{{\cos }^3}\left( {\dfrac{{\theta - \phi }}{2}} \right) - 3\cos \left( {\dfrac{{\theta - \phi }}{2}} \right)} \right]}}{{2\cos (\theta - \phi ) - 1}}............(3)\]

We know the formula for cos2x, given as follows:

\[\cos 2x = 2{\cos ^2}x - 1\]

Using this formula in equation (3) for \[2\cos (\theta - \phi )\], we get:
\[LHS = \dfrac{{2\cos \left( {3\left( {\dfrac{{\theta + \phi }}{2}} \right)} \right)\left[ {4{{\cos }^3}\left( {\dfrac{{\theta - \phi }}{2}} \right) - 3\cos \left( {\dfrac{{\theta - \phi }}{2}} \right)} \right]}}{{4{{\cos }^2}\left( {\dfrac{{\theta - \phi }}{2}} \right) - 3}}............(4)\]

We can cancel few terms in equation (4) to obtain the equation as follows:

\[LHS = 2\cos \left( {\dfrac{{3\theta + 3\phi }}{2}} \right)\cos \left( {\dfrac{{\theta - \phi }}{2}} \right).........(5)\]

Hence, we simplified the LHS to equation (5).

We now simplify the right-hand side by assigning it to RHS.

\[RHS = (\cos \theta + \cos \phi )\cos (\theta + \phi ) - (\sin \theta + \sin \phi )\sin (\theta + \phi )\]

We multiply the terms and try to simplify the following:

\[RHS = \cos \theta \cos (\theta + \phi ) + \cos \phi \cos (\theta + \phi ) - \sin \theta \sin (\theta + \phi ) - \sin \phi \sin (\theta + \phi )......(6)\]

We can rewrite equation (6) as follows:

\[RHS = \cos \theta \cos (\theta + \phi ) - \sin \theta \sin (\theta + \phi ) + \cos \phi \cos (\theta + \phi ) - \sin \phi \sin (\theta + \phi )......(7)\]

We know the following formula:

\[\cos A\cos B - \sin A\sin B = \cos (A + B).......(8)\]

Using formula (8) in equation (7), we get:

\[RHS = \cos (2\theta + \phi ) + \cos (\theta + 2\phi ).........(9)\]

Now, we use the formula of sum of cosines as follows:

\[\cos A + \cos B = 2\cos \dfrac{{A + B}}{2}\cos \dfrac{{A - B}}{2}\]

\[RHS = 2\cos \dfrac{{(\theta + 2\phi ) + (\theta + 2\phi )}}{2}\cos \dfrac{{(2\theta + \phi ) - (\theta + 2\phi )}}{2}\]

Simplifying this equation, we get:

\[RHS = 2\cos \left( {\dfrac{{3\theta + 3\phi }}{2}} \right)\cos \left( {\dfrac{{\theta - \phi }}{2}} \right)..........(10)\]

Comparing equation (5) with equation (10), we have:

LHS=RHS

Hence, proved.


Note: It will be difficult to directly go from the left-hand side of the equation to the right-hand side. So, simplify both sides to a common term and hence, prove by equality.