Algebra 2

1st Edition

ISBN: 9780078884825

Author: McGraw-Hill/Glencoe

Publisher: Glencoe/McGraw-Hill School Pub Co

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Question

Chapter 14.2, Problem 59HP

To determine

To Justify : The three basic Pythagorean Identities.

Expert Solution & Answer

Explanation of Solution

Given information :

We know that, Pythagorean Identities.

Formula used :

$(hypotenuse)2=(perpendicular)2+(base)2$

Proof :

$In triangle ABC, b 2 = c 2 + a 2 ....(1) (Using Pythagoras theorem)$

$Dividing by b 2$

$b 2 b 2 = c 2 b 2 + a 2 b 2 ⇒ 1= c 2 b 2 + a 2 b 2$

$Now, sin θ= AB AC = c b , cosθ= BC AC = a b$

$1= sin 2 θ+ cos 2 θ$

$sin 2 θ+ cos 2 θ=1 ......(2)$

$Dividing (1) by c 2$

$b 2 c 2 = c 2 c 2 + a 2 c 2$

$b 2 c 2 =1+ a 2 c 2$

$cosec 2 θ= AC AB = b c , cotθ= BC AB = a c$

$cosec 2 θ=1+ cot 2 θ$

$cosec 2 θ− cot 2 θ=1 .....(3)$

$Dividing (1) by a 2$

$b 2 a 2 = c 2 a 2 + a 2 a 2$

$tanθ= AB BC = c a , secθ= AC BC = b a$

$sec 2 θ= tan 2 θ+1$

$sec 2 θ− tan 2 θ=1 .....(4)$

$Fromm (2), (3) and (4) Pythgorean identities are verified.$

$sin 2 θ+ cos 2 θ=1$

$cosec 2 θ− cot 2 θ=1$

$sec 2 θ− tan 2 θ=1$

Algebra 2, Chapter 14.2, Problem 59HP

Chapter 14.2, Problem 58HP

Chapter 14.2, Problem 60STP

Chapter 14 Solutions

Algebra 2

Ch. 14.1 - Prob. 1ACYP Ch. 14.1 - Prob. 1BCYP Ch. 14.1 - Prob. 2ACYP Ch. 14.1 - Prob. 2BCYP Ch. 14.1 - Prob. 3CYP Ch. 14.1 - Prob. 1CYU Ch. 14.1 - Prob. 2CYU Ch. 14.1 - Prob. 3CYU Ch. 14.1 - Prob. 4CYU Ch. 14.1 - Prob. 5CYU

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