Question

If a + b = c then ${\cos ^2}A + {\cos ^2}B + {\cos ^2}C - 2\cos A\cos B\cos C$ is equal to
A) $1$
B) $2$
C) $0$
D) $3$

Answer 1

We will take cos on both the sides of a + b = c. We will use the trigonometric identity $\cos 2A = 2{\cos ^2}A - 1$. We know that $\cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})$. We will use all these formulas to find the value and get the final output. Trigonometry can be divided into two sub-branches called plane trigonometry and spherical geometry.

Complete step by step answer:
Given that,
A + B = C
Taking cos on both the sides, we get,
$\cos (A + B) = \cos C$ ---- (i)
Similarly, taking sin on both the sides, we get,
$\sin (A + B) = \sin C$
We know that,
$\cos 2A = 2{\cos ^2}A - 1$
$\Rightarrow {\cos ^2}A = \dfrac{{\cos 2A + 1}}{2}$ -------- (ii)
We need to find the value of
${\cos ^2}A + {\cos ^2}B + {\cos ^2}C - 2\cos A\cos B\cos C$
Substituting the value of (ii) in the above expression, we will get,
$= \dfrac{{\cos 2A + 1}}{2} + \dfrac{{\cos 2B + 1}}{2} + \dfrac{{\cos 2C + 1}}{2} - 2\cos A\cos B\cos C$
On evaluating this, we get,
$= \dfrac{{\cos 2A + 1 + \cos 2B + 1 + \cos 2C + 1}}{2} - 2\cos A\cos B\cos C$
$= \dfrac{{3 + \cos 2A + \cos 2B + \cos 2C}}{2} - 2\cos A\cos B\cos C$
$= \dfrac{3}{2} + \dfrac{{\cos 2A + \cos 2B + \cos 2C}}{2} - 2\cos A\cos B\cos C$
$= \dfrac{3}{2} + \dfrac{1}{2}(\cos 2A + \cos 2B + \cos 2C) - 2\cos A\cos B\cos C$
$= \dfrac{3}{2} + \dfrac{1}{2}((\cos 2A + \cos 2B) + \cos 2C) - 2\cos A\cos B\cos C$
We know that, $\cos C + \cos D = 2\cos (\dfrac{{C + D}}{2})\cos (\dfrac{{C - D}}{2})$
$= \dfrac{3}{2} + \dfrac{1}{2}(2\cos (A + B)\cos (A - B)) + \cos 2C) - 2\cos A\cos B\cos C$
We know that, $2\cos A\cos B = \cos (A + B) + \cos (A - B)$
$= \dfrac{3}{2} + \dfrac{1}{2}(2\cos (A + B)\cos (A - B) + \cos 2C) - (\cos (A + B) + \cos (A - B))\cos C$
Substituting the value from (i) in the above expression, we get,
$= \dfrac{3}{2} + \dfrac{1}{2}(2\cos C\cos (A - B) + \cos 2C) - (\cos C + \cos (A - B))\cos C$
Removing the brackets, we get,
$= \dfrac{3}{2} + \cos C\cos (A - B) + \dfrac{{\cos 2C}}{2} - {\cos ^2}C + \cos (A - B)(\cos C)$
Simplify this expression, we will get,
$= \dfrac{3}{2} + \dfrac{{\cos 2C}}{2} - {\cos ^2}C$
$= \dfrac{3}{2} + \dfrac{{\cos 2C - 2{{\cos }^2}C}}{2}$
We know that, $\cos 2A = 2{\cos ^2}A - 1$
$= \dfrac{3}{2} + \dfrac{{2{{\cos }^2}C - 1 - 2{{\cos }^2}C}}{2}$
$= \dfrac{3}{2} - \dfrac{1}{2}$

$$= \dfrac{{3 - 1}}{2} \\\ = \dfrac{2}{2} \\\ = 1 \\\$$

Hence, if a + b = c is given then the value of ${\cos ^2}A + {\cos ^2}B + {\cos ^2}C - 2\cos A\cos B\cos C = 1$.

Note:
Trigonometry helps us to understand the relation between the sides and angles of a right-angle triangle. Cosec, Sec and Cot are three of the six trigonometric ratios of a right-angled triangle. The basics of trigonometry define three primary functions which are sine, cosine and tangent. The trigonometric ratios of a triangle are also called the trigonometric functions. The angles are either measured in radians or degrees. The first trigonometric table was apparently compiled by Hipparchus, who is consequently now known as "the father of trigonometry".