Trigonometry 7th Edition

Published by Cengage Learning
ISBN 10: 1111826854
ISBN 13: 978-1-11182-685-7

Chapter 1 - Section 1.3 - Definition I: Trigonometric Functions - 1.3 Problem Set - Page 32: 30

Answer

The point is $(-1,-1)$ $r=\sqrt{2}$ $\sin({-135^{\circ}}) = -\dfrac{\sqrt{2}}{2}$ $\cos({-135}^{\circ}) = -\dfrac{\sqrt{2}}{2}$ $\tan({-135}^{\circ}) = 1$
1536411731

Work Step by Step

The terminal side of $-135^{\circ}$ in standard position is represented by the blue line in the figure. It lies in the 3rd quadrant. The coordinates of points on the terminal side of $-135^{\circ}$ can be given by $(-a,-a)$, where $a$ is a positive number. Choosing $a=1$ arbitrarily, the point is $(-1,-1)$. To find the distance from the origin to $(-1,-1)$, we use the distance formula $$r=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\r=\sqrt{(-1-0)^2+(-1-0)^2}=\sqrt{2}$$ $$\therefore r = \boxed{\sqrt{2}}$$ $\sin({-135^{\circ}}) = \dfrac{y}{r} = \dfrac{-1}{\sqrt{2}} = \boxed{-\dfrac{\sqrt{2}}{2}}$ $\cos({-135}^{\circ}) = \dfrac{x}{r} = \dfrac{-1}{\sqrt{2}} = \boxed{-\dfrac{\sqrt{2}}{2}} $ $\tan({-135}^{\circ}) = \dfrac{y}{x} = \dfrac{-1}{-1 } = \boxed{1}$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.