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.,
$$\sin A \pm \sin B = 2\sin \left( {\dfrac{{A \pm B}}{2}} \right)\cos \left( {\dfrac{{A \mp B}}{2}} \right)$$
$$\cos A - \cos B = - 2\sin \left( {\dfrac{{A + B}}{2}} \right)\sin \left( {\dfrac{{A - B}}{2}} \right)$$
$$\cos A + \cos B = 2\cos \left( {\dfrac{{A + B}}{2}} \right)\cos \left( {\dfrac{{A - B}}{2}} \right)$$