Chapter 8.2, Problem 58AYU

To determine

To find: Derive the states Mollwiede’s formula for any triangle.

Answer to Problem 58AYU

The Mollweide’s formula is

  $\frac{a+b}{c}=\frac{\cos \left[\frac{\left(A-B\right)}{2}\right]}{\sin \left(\frac{C}{2}\right)}$ ,

Explanation of Solution

Given information:

The given triangle is ABC

Its sides and opposite angle are $a,b,c$ ,

Concept used:

Law of Sine:

If ABC is a triangle with sides $a$ , $b$ and $c$ , then

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

The relation of between the sides of triangle ABC is.

  $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

By using the law of Sine,

  $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

Multiplying with $a$ of both sides to find the values of $\frac{a}{c}$

  $\begin{array}{l}\left(\frac{\sin A}{a}\right)\times a=\left(\frac{\sin C}{c}\right)\times a\\ \sin A=\frac{a}{c}\times \sin C\\ \frac{a}{c}=\frac{\sin A}{\sin C}\mathrm{....}\left(1\right)\end{array}$

To find the values of $\frac{b}{c}$ and $\frac{a}{b}$ s

From equation $\mathrm{....}\left(1\right)$

  $\frac{b}{c}=\frac{\sin B}{\sin C}\mathrm{....}\left(2\right)$

  $\frac{a}{b}=\frac{\sin A}{\sin B}\mathrm{.....}\left(3\right)$

The relation of between the sides of triangle ABC is.

  $\frac{a+b}{c}=\frac{a}{c}+\frac{b}{c}$

Substitute the values of equation $\mathrm{....}\left(1\right)$ and $\mathrm{....}\left(2\right)$

  $\frac{\sin A}{\sin C}+\frac{\sin B}{\sin C}=\frac{\sin A+\sin B}{\sin C}\mathrm{......}\left(4\right)$

Simplify by the sum to product formula,

  $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)$

Simplify by half angle formula,

  $\sin C=2\sin \frac{c}{2}.\cos \frac{c}{2}$

To put the above values in equation $\mathrm{.....}\left(4\right)$

  $\begin{array}{c}\frac{\sin A+\sin B}{\sin C}=\frac{2\sin \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)}{2\sin \frac{C}{2}\cos \frac{C}{2}}\\ =\frac{\sin \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}.\cos \frac{C}{2}}\mathrm{....}\left(5\right)\end{array}$

The sum of angle of the triangle $ABC$ is,

  $\begin{array}{l}A+B+C=\pi \\ A+B=\pi -C\end{array}$

Substitute the values of $A+B=\pi -C$ in equation $\mathrm{.....}\left(5\right)$

  $\begin{array}{c}\frac{\sin \left(\frac{A+B}{2}\right).\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}.\cos \frac{C}{2}}=\frac{\sin \left(\frac{\pi -C}{2}\right).\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}.\cos \frac{C}{2}}\\ =\frac{\sin \left(\frac{\pi }{2}-\frac{C}{2}\right).\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}.\cos \frac{C}{2}}\end{array}$

The angle of $\sin \left(\frac{\pi }{2}-\theta \right)=\cos \theta$

Applied it $\sin \left(\frac{\pi }{2}-\frac{C}{2}\right)=\cos \frac{C}{2}$

  $\begin{array}{c}\frac{\sin \left(\frac{\pi }{2}-\frac{C}{2}\right).\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}.\cos \frac{C}{2}}=\frac{\cos \frac{C}{2}.\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}.\cos \frac{C}{2}}\\ =\frac{\cos \left(\frac{A-B}{2}\right)}{\sin \frac{C}{2}}\end{array}$

Hence, the derive states of the Mollweide’s formula is $\frac{a+b}{c}=\frac{\cos \left[\frac{\left(A-B\right)}{2}\right]}{\sin \left(\frac{C}{2}\right)}$ .