Question

In a triangle $ABC$, $\cos mA + \cos mB + \cos mC - 1 = \pm 4\sin \dfrac{{mA}}{2}\sin \dfrac{{mB}}{2}\sin \dfrac{{mC}}{2}$ according as $m$ is of the form $4n + 1$ or $4n + 3$is.

A.True

B.False

Answer 1

We know that in a triangle $ABC$ the sum of all the angles is equal to $180^\circ$ that is $A + B + C = 180^\circ$ or $\dfrac{{A + B}}{2} = 90 - \dfrac{C}{2}$. We will use the property of $\cos A$ is equal to $1 - 2{\sin ^2}\left( {\dfrac{C}{2}} \right)$. Also, we have to use the property that $\cos A + \cos B$ is equal to $2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$.

Complete step by step solution
Given:
It is given that $ABC$is a triangle.

So, we have $A + B + C = 180^\circ$, since the sum of angles of the triangle is $180^\circ$.

Our aim is to show that,
$\cos mA + \cos mB + \cos mC - 1 = \pm 4\sin \dfrac{{mA}}{2}\sin \dfrac{{mB}}{2}\sin \dfrac{{mC}}{2}$......(1)

Since, we can write the above equation in the form of,$\cos mA + \cos mB + \cos mC = 1 \pm 4\sin \dfrac{{mA}}{2}\sin \dfrac{{mB}}{2}\sin \dfrac{{mC}}{2}$......(2)

We will solve further by taking the left hand side and show that it is equal to the right hand side.

Now let us take $\cos mA + \cos mB + \cos mC$.

We will now use the formula that is,
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$

Now, on applying the above formula in $\cos mA + \cos mB + \cos mC$ we get,
$\begin{array}{l} 2\cos \left( {\dfrac{{mA + mB}}{2}} \right)\cos \left( {\dfrac{{mA - mB}}{2}} \right) + \cos mC\\\ 2\cos \left( {\dfrac{{m\left( {A + B} \right)}}{2}} \right)\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) + \cos mC \end{array}$

Also, let us use another formula of trigonometry that is,
$\cos 2C = 1 - 2{\sin ^2}C$

On applying the above formula in $2\cos \left( {\dfrac{{m\left( {A + B} \right)}}{2}} \right)\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) + \cos mC$ we get,
$2\cos \left( {\dfrac{{m\left( {A + B} \right)}}{2}} \right)\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{{mC}}{2}} \right)$

Since it is known that $ABC$ is triangle, so we get,
$\begin{array}{l} A + B = 180 - C\\\ \dfrac{{A + B}}{2} = 90 - \dfrac{C}{2} \end{array}$

So now let us use this in the above equation, we get,
$\begin{array}{l} 2\cos \left( {m\left( {90 - \dfrac{C}{2}} \right)} \right)\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{{mC}}{2}} \right)\\\ 2\cos \left( {90m - \dfrac{C}{2}} \right)\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) + 1 - {\sin ^2}\left( {\dfrac{{mC}}{2}} \right) \end{array}$

Now, we will use the formula from the trigonometry, we have,
$\cos \left( {m90 - \dfrac{{mC}}{2}} \right) = \pm \sin \left( {\dfrac{{mC}}{2}} \right)$

Where, $m$ is of the form $4n + 1$, $4n + 3$.

On Substituting the above equation in $2\cos \left( {90m - \dfrac{{mC}}{2}} \right)\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) + 1 - {\sin ^2}\left( {\dfrac{{mC}}{2}} \right)$ we get,
$\pm 2\sin \left( {\dfrac{{mC}}{2}} \right)\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) + 1 - {\sin ^2}\left( {\dfrac{{mC}}{2}} \right)$

Since, $\sin \left( {\dfrac{{mC}}{2}} \right)$ is common in both let us take common.
$\pm 2\sin \left( {\dfrac{{mC}}{2}} \right)\left( {\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) - \sin \left( {\dfrac{{mC}}{2}} \right)} \right) + 1$

Since, we know that $\dfrac{C}{2} = \dfrac{{A + B}}{2}$. So, we will substitute in the above equation we get,
$\pm 2\sin \left( {\dfrac{{mC}}{2}} \right)\left( {\cos \left( {\dfrac{{m\left( {A - B} \right)}}{2}} \right) - \sin \left( {\dfrac{{m\left( {A + B} \right)}}{2}} \right)} \right) + 1$

Let us use the formula that is $\cos \left( A \right) - \cos \left( B \right) = 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$,

Here, in our problem $A = A - B$ and $B = A + B$ so, we get,

$\begin{array}{l} \pm 2\sin \left( {\dfrac{{mC}}{2}} \right)\left( {2\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)} \right) + 1\\\ \pm 4\sin \left( {\dfrac{{mC}}{2}} \right)\left( {\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)} \right) + 1 \end{array}$

Hence, the right hand side is proved that is,$\cos mA + \cos mB + \cos mC = 1 \pm 4\sin \dfrac{{mA}}{2}\sin \dfrac{{mB}}{2}\sin \dfrac{{mC}}{2}$.

Here, we can see that $m$ is of the form $4n + 1$ or $4n + 3$ we get,

$\cos mA + \cos mB + \cos mC - 1 = \pm 4\sin \dfrac{{mA}}{2}\sin \dfrac{{mB}}{2}\sin \dfrac{{mC}}{2}$.

Therefore, the answer is true.

Note: We are not taking $m$ as even because $\cos \left( {m90 - \dfrac{{mC}}{2}} \right)$. For example take $m = 2$ then we get $\cos \left( {180 - \dfrac{{mC}}{2}} \right) = \cos \left( {\dfrac{{mC}}{2}} \right)$ . Here, we got $\cos$ term but in the proof all the terms should be in $\sin$. We will not get the $\sin$ term if we take $m$ as $4n$ or $4n + 2$. So the only possibilities for $m$ are $4n + 1$ and $4n + 3$.