Question

If A, B, C are the angles of a triangle then $\cos A + \cos B + \cos C = ?$

Answer 1

Hint : Here in this question, we have to find the value given trigonometric expression. For this, first we need to consider the sum of inner angles of triangle $\vartriangle ABC$, then apply a double or half angle formula, sum to product formula and other standard formula on simplification we get the required solution.

Complete step by step solution:
A function of an angle expressed as the ratio of two of the sides of a right triangle that contains that angle; the sine, cosine, tangent, cotangent, secant, or cosecant known as trigonometric function Also called circular function.
Consider the given question:
If A, B, C are angles of a triangle, then we know the sum of all interior angles of a triangle should be equal to ${180^ \circ }$.
$\Rightarrow \,\,A + B + C = {180^ \circ }$
And
$\Rightarrow \,\,A + B = {180^ \circ } - C$.
Consider,
$\Rightarrow \,\,\,\cos A + \cos B + \cos C$ --------- (1)
Let us by the sum to product formula and double angle of trigonometric ratios:
The sum to product of cosine ratio is:
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right) \cdot \cos \left( {\dfrac{{A - B}}{2}} \right)$.
The double angle of cosine ratio is:
$\cos 2A = 1 - 2{\sin ^2}A$ or $\cos A = 1 - 2{\sin ^2}\left( {\dfrac{A}{2}} \right)$.
Then equation (1) becomes
$\Rightarrow \,\,\,2\cos \left( {\dfrac{{A + B}}{2}} \right) \cdot \cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{C}{2}} \right)$
But $A + B = {180^ \circ } - C$
$\Rightarrow \,\,\,2\cos \left( {\dfrac{{{{180}^ \circ } - C}}{2}} \right) \cdot \cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{C}{2}} \right)$
$\Rightarrow \,\,\,2\cos \left( {\dfrac{{{{180}^ \circ }}}{2} - \dfrac{C}{2}} \right) \cdot \cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{C}{2}} \right)$
$\Rightarrow \,\,\,2\cos \left( {{{90}^ \circ } - \dfrac{C}{2}} \right) \cdot \cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{C}{2}} \right)$ ----- (2)
By the ASTC rule $\left( {{{90}^ \circ } - \theta } \right)$ belongs to the first quadrant in that all six ratios are positive and
Let us by the complementary angles of trigonometric ratios:
The angle can be written as
$\sin \left( {90 - \theta } \right) = \cos \theta$
$\cos \left( {90 - \theta } \right) = \sin \theta$
Then equation (2) becomes
$\Rightarrow \,\,\,2\sin \left( {\dfrac{C}{2}} \right) \cdot \cos \left( {\dfrac{{A - B}}{2}} \right) + 1 - 2{\sin ^2}\left( {\dfrac{C}{2}} \right)$
Take $2\sin \left( {\dfrac{C}{2}} \right)$ as common term, then we have
$\Rightarrow \,\,\,2\sin \left( {\dfrac{C}{2}} \right)\left( {\cos \left( {\dfrac{{A - B}}{2}} \right) - \sin \left( {\dfrac{C}{2}} \right)} \right) + 1$
We already showed $\cos \left( {\dfrac{{A + B}}{2}} \right) = \sin \left( {\dfrac{C}{2}} \right)$, then
$\Rightarrow \,\,\,2\sin \left( {\dfrac{C}{2}} \right)\left( {\cos \left( {\dfrac{{A - B}}{2}} \right) - \cos \left( {\dfrac{{A + B}}{2}} \right)} \right) + 1$ ---- (3)
Again by the sum to product formula
$\cos A - \operatorname{Cos} B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right) \cdot \sin \left( {\dfrac{{A - B}}{2}} \right)$
Or
$\cos \left( {\dfrac{{A + B}}{2}} \right) - \operatorname{Cos} \left( {\dfrac{{A + B}}{2}} \right) = - 2\sin \left( {\dfrac{A}{2}} \right) \cdot \sin \left( {\dfrac{B}{2}} \right)$
Then equation (3) becomes
$\Rightarrow \,\,\,2\sin \left( {\dfrac{C}{2}} \right)\left( {2\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)} \right) + 1$
Then on simplification, we get
$\therefore \,\,\,4\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right) + 1$
Hence, the required value is
So, the correct answer is “$\,\,\,4\sin \left( {\dfrac{A}{2}} \right)\sin \left( {\dfrac{B}{2}} \right)\sin \left( {\dfrac{C}{2}} \right) + 1$”.

Note : When solving the trigonometry-based questions, we have to know the definitions of six trigonometric ratios. Remember, the sum to product formula of trigonometry i.e.,
$\sin A \pm \sin B = 2\sin \left( {\dfrac{{A \pm B}}{2}} \right)\cos \left( {\dfrac{{A \mp B}}{2}} \right)$
$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$
$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$