Question

The half-angle identitytan A/2 = ± √[(1 - cosA)/(1 + cos A)]can be used to find tan 22.5° = √(3 - 2√2), and the half-angle identitytan A/2 = sin A/(1 + cos A)can be used to find tan 22.5° = √2 - 1. Show that these answers are the same, without using a calculator. (Hint: If a \u003e 0 and b \u003e 0 and a² = b², then a = b.)

Accepted Answer

Start by writing down the two expressions for $$ 	an 22.5^\circ $$ given in the problem: $$ 	an 22.5^\circ = \sqrt{3 - 2\sqrt{2}} $$ and $$ 	an 22.5^\circ = \sqrt{2} - 1 . $$Our goal is to show these two expressions are equal without using a calculator. Square both expressions to use the hint that if $$ a \u003e 0 $$, $$ b \u003e 0 $$, and $$ a^2 = b^2 $$, then $$ a = b . $$First, square $$ \sqrt{3 - 2\sqrt{2}}  to $$get $$ 3 - 2\sqrt{2} . $$Next, square $$ \sqrt{2} - 1 . $$Use the formula for squaring a binomial: $$ (x - y)^2 = x^2 - 2xy + y^2 . So$$, $$ (\sqrt{2} - 1)^2 = (\sqrt{2})^2 - 2 	imes \sqrt{2} 	imes 1 + 1^2 = 2 - 2\sqrt{2} + 1 . $$Simplify the squared form of $$ \sqrt{2} - 1  to $$get $$ 3 - 2\sqrt{2} $$, which matches the squared form of the first expression. Since both squared expressions are equal and both original expressions are positive (as tangent of 22.5° is positive), conclude that $$ \sqrt{3 - 2\sqrt{2}} = \sqrt{2} - 1 $$, proving the two forms of $$ 	an 22.5^\circ $$ are the same.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Half-Angle Identities for Tangent

Simplifying Radical Expressions

Using the Property of Equality of Squares