Trigonometry (11th Edition) Clone

Published by Pearson
ISBN 10: 978-0-13-421743-7
ISBN 13: 978-0-13421-743-7

Chapter 5 - Trigonometric Identities - Section 5.6 Half-Angle Identities - 5.6 Exercises - Page 242: 18


$$\sqrt{3-2\sqrt 2}=\sqrt 2-1$$ The equation is proved. 2 results are the same.

Work Step by Step

Prove that $$\sqrt{3-2\sqrt 2}=\sqrt 2-1$$ *Method 1: Transform the left side We see that the left side has a giant square root which the right side does not have. So it would be better trying to eliminate it to see that two sides are equal or not. To do so, we can use $$\sqrt{x^2}=|x|$$ Therefore, the job now is to figure out a way to change $3-2\sqrt2$ into a squared formula, which is quite straightforward after changing a little bit: $$3-2\sqrt2$$ $$=2-2\times\sqrt 2\times1+1$$ $$=(\sqrt 2)^2-2\times\sqrt 2\times1+1^2$$ $$=(\sqrt 2-1)^2$$ Therefore, applying it to the left side, we have $$\sqrt{3-2\sqrt 2}$$ $$=\sqrt{(\sqrt 2-1)^2}$$ $$=|\sqrt 2-1|$$ $$=\sqrt 2-1$$ (since $\sqrt 2-1\gt0$) That means the equation has been proved. 2 results are the same. *Method 2: Using the textbook hints, applying the formula If $a\gt0$ and $b\gt0$ and $a^2=b^2$, then $a=b$ We see that $3\gt2\sqrt 2$, so $3-2\sqrt 2\gt0$, which means $\sqrt{3-2\sqrt 2}\gt0$ $\sqrt2\gt1$, so $\sqrt 2-1\gt0$ Now we take the square of both sides: - Left side: $$(\sqrt{3-2\sqrt 2})^2=3-2\sqrt 2$$ - Right side: $$(\sqrt 2-1)^2=2+1-2\sqrt 2=3-2\sqrt 2$$ That shows $(\sqrt{3-2\sqrt 2})^2=(\sqrt 2-1)^2$ Therefore, $\sqrt{3-2\sqrt2}=\sqrt 2-1$
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