Question

If \[\sin A - \sin B = c\] and \[\cos A - \cos B = d\] then find \[\cos e{c^2}[\dfrac{{(A + B)}}{2}]\].

Answer 1

Hint: In order to solve the given question, we have to expand the equation with help of theorem and apply division rule.
The question requires application of trigonometry rules to arrive at the solution. Trigonometry is a branch of mathematics that investigates the relationship between triangle side lengths and angles.

Complete step-by-step answer:
First, we apply the formula of \[\sin A - \sin B\]:
\[\sin A - \sin B = 2\sin (\dfrac{{A - B}}{2})\cos (\dfrac{{A + B}}{2})\]
Since we are given that \[\sin A - \sin B\]is equal to \[c\], comparing with the above formula, we get,
\[2\sin (\dfrac{{A - B}}{2})\cos (\dfrac{{A + B}}{2}) = c\] …(1)
Similarly, we apply the formula of \[\cos A - \cos B\]:
\[\cos A - \cos B = - 2\sin (\dfrac{{A - B}}{2})\sin (\dfrac{{A + B}}{2})\]
Since we are given that \[\cos A - \cos B\] is equal to \[d\], comparing with the above formula, we get,
\[ - 2\sin (\dfrac{{A - B}}{2})\sin (\dfrac{{A + B}}{2}) = d\] …(2)
Now in order to find \[\cos ec\], we have to divide equation (1) by equation (2) as follows:
\[ = \dfrac{{2\sin (\dfrac{{A - B}}{2})\cos (\dfrac{{A + B}}{2})}}{{\left( { - 2} \right)\sin (\dfrac{{A - B}}{2})\sin (\dfrac{{A + B}}{2})}} = \dfrac{c}{d}\]
Cancelling the common terms, we get,
\[ = ( - 1)\dfrac{{\cos (\dfrac{{A + B}}{2})}}{{\sin (\dfrac{{A + B}}{2})}} = \dfrac{c}{d}\]
Now we know that \[\dfrac{{\sin \theta }}{{\cos \theta }} = \cot \theta \], hence comparing with the above equation, we get,
\[( - 1)\cot (\dfrac{{A + B}}{2}) = \dfrac{c}{d}\]

Now taking negative sign on the other side of equation, we get,
\[\cot (\dfrac{{A + B}}{2}) = - \dfrac{c}{d}\]
Now as per the theorem, we know that:
\[\cos e{c^2}\theta = 1 + {\cot ^2}\theta \]
So, comparing with above formula, we get,
\[\cos e{c^2}(\dfrac{{A + B}}{2}) = 1 + {\cot ^2}(\dfrac{{A + B}}{2})\]
Substituting the value of \[\cot (\dfrac{{A + B}}{2}) = - \dfrac{c}{d}\],
\[\cos e{c^2}(\dfrac{{A + B}}{2}) = 1 + {( - \dfrac{c}{d})^2}\]
Squaring the negative bracket on RHS, we get,
\[\cos e{c^2}(\dfrac{{A + B}}{2}) = 1 + \dfrac{{{c^2}}}{{{d^2}}}\]
Hence, after cross multiplication we will get,
\[\cos e{c^2}(\dfrac{{A + B}}{2}) = \dfrac{{{c^2} + {d^2}}}{{{d^2}}}\]

Note: Here \[\dfrac{{\sin \theta }}{{\cos \theta }} = \cot \theta \]and \[\cos e{c^2}\theta = 1 + {\cot ^2}\theta \]are trigonometric identities based on theorems. A theorem is a statement that has been proven true by a rigorous proof with help of logical argument. We know with 100 percent certainty that a theorem is valid because it has been proven. There are different trigonometric identities which can be used at arriving the solution. However, the final answer will remain the same.