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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 8: Q. 57 (page 537)

For any triangle show that $\sin \frac{C}{2} = \sqrt{\frac{(s - a) (s - b)}{a b}}$ where $s = \frac{1}{2} (a + b + c)$

Short Answer

Expert verified

Hence the result $\sin \frac{C}{2} = \sqrt{\frac{(s - a) (s - b)}{a b}}$ has been proved.

Step by step solution

01

Step 1. Given Information

$s = \frac{1}{2} (a + b + c)$

02

Step 2. Use half-angle formula

By using the half-angle formula,

$\cos C = 1 - 2 \sin^{2} \frac{C}{2}$

From the equation,

$\sin^{2} \frac{C}{2} = \frac{1 - \cos C}{2} \sin \frac{C}{2} = \sqrt{\frac{1 - \cos C}{2}}$

03

Step 3. The law of cosines.

Let us consider a triangle having sides $a, b, c$ and the corresponding opposite angles $A, B, C$ respectively.

By the law of cosines,

$\cos C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}$

04

Step 4. Solve the equation

Substitute the value of $\cos C$ in the first equation and solve,

$\sqrt{\frac{1 - \cos C}{2}} = \sqrt{\frac{1 - \sqrt{\frac{a^{2} + b^{2} - c^{2}}{2 a b}}}{2}} = \sqrt{\frac{- (a^{2} + b^{2} - 2 a b) + c^{2}}{4 a b}} = \sqrt{\frac{c^{2} - {(a - b)}^{2}}{4 a b}}$

05

Step 5. Solve the numerator.

$\sqrt{\frac{c^{2} - {(a - b)}^{2}}{4 a b}} = \sqrt{\frac{(c + a - b) (c - a - b)}{4 a b}}$

Since

$s = \frac{1}{2} (a + b + c) 2 s = a + b + c c = 2 s - a - b$

06

Step 5. Substitute for c

$\sqrt{\frac{(c + a - b) (c - a - b)}{4 a b}} = \sqrt{\frac{(2 s - a - b + a - b) (2 s - a - b - a - b)}{4 a b}} = \sqrt{\frac{(2 s - 2 b) (2 s - 2 a)}{4 a b}} = \sqrt{\frac{(s - b) (s - a)}{a b}}$

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Most popular questions from this chapter

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