Chapter 6.6, Problem 38E

To determine

Prove the given equations.

Answer to Problem 38E

Result is proved to be true.

Explanation of Solution

Given information:

Given,

  $\begin{array}{l}a=b\cos C+c\cos B\\ b=c\cos A+a\cos C\\ c=a\cos B+b\cos A\end{array}$

These are called projection laws.

Calculation:

We need to prove that in triangle $ABC$ ,

  $\begin{array}{l}a=b\cos C+c\cos B\\ b=c\cos A+a\cos C\\ c=a\cos B+b\cos A\end{array}$

We will use the laws of cosines to prove the above identities

For any triangle $ABC$ , the law of cosines is given by

  $\begin{array}{l}{a}^2=b^2+c^2-2bc\cos A\\ b^2=c^2+c^2-2ac\cos B\\ c^2=a^2+b^2-2ab\cos C\end{array}$

Adding the equations of law of cosines for angles $BandC$ , we get,

  $b^2+c^2=2a^2+b^2+c^2-2a(b\cos C+c\cos B)$

Cancelling $b^2+c^2$ , we get,

  $0=2a^2-2a(b\cos C+c\cos B)$

Dividing the above equation by $a$ , we get,

  $0=a^{}-(b\cos C+c\cos B)$

Solving for $a$ ,

So,

  $a=b\cos C+c\cos B$

Again, adding the equations of law of cosines for angles $AandC$ ,we get,

  $a^2+c^2=a^2+c^2+2b^2-2a(c\cos A+a\cos C)$

Cancelling $a^2+c^2$ , we get,

  $0=2b^2-2b(c\cos A+a\cos C)$

Dividing the above equation by $a$ , we get,

  $0=b-(c\cos A+a\cos C)$

Solving for $b$ ,

So,

  $b=c\cos A+a\cos C$

Again, adding the equations of law of cosines for angles $AandB$ , we get,

  $a^2+b^2=a^2+b^2+2c^2-2c(a\cos B+b\cos A)$

Cancelling $a^2+b^2$ , we get,

  $0=2c^2-2c(a\cos B+b\cos A)$

Dividing the above equation by $a$ , we get,

  $0=c-(a\cos B+b\cos A)$

Solving for $c$ ,

So,

  $c=a\cos B+b\cos A$

Hence, it is proved above.