Answered: Evaluate the sine, cosine, and tangent…

Name: Evaluate the sine, cosine, and tangent of each angle.
Uploaded: 2024-03-20T06:46:55.000Z
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Description: Evaluate the sine, cosine, and tangent of each angle. (a) 225° (b) −225° (c) 5/3 (d) 11/6

Holt Mcdougal Larson Pre-algebra: Student Edition 2012

1st Edition

ISBN:9780547587776

Author:HOLT MCDOUGAL

Publisher:HOLT MCDOUGAL

Chapter10: Measurement, Area, And Volume

Section10.1: Triangles

Problem 29E

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Evaluate the sine, cosine, and tangent of each angle. (a) 225° (b) −225° (c) 5/3 (d) 11/6

Expert Solution

Step 1

Evaluate sine, cosine and tangent.

Step 2

a) When $θ = 225 °$ :

$\begin{array}{rcl} \sin (225 °) & = & \sin (180 ° + 45 °) \\ = & - \sin (45 °) [S i n c e \sin (180 ° + θ) = - \sin θ] \\ = & - \frac{1}{\sqrt{2}} \\ \cos (225 °) & = & \cos (180 ° + 45 °) \\ = & - \cos (45 °) [S i n c e \cos (180 ° + θ) = - \cos θ] \\ = & - \frac{1}{\sqrt{2}} \\ \tan (225 °) & = & \tan (180 ° + 45 °) \\ = & \tan (45 °) [S i n c e \tan (180 ° + θ) = \tan θ] \\ = & 1 \end{array}$

Hence

$\begin{array}{rcl} \sin (225 °) & = & - \frac{1}{\sqrt{2}} \\ \cos (225 °) & = & - \frac{1}{\sqrt{2}} \\ \tan (225 °) & = & 1 \end{array}$

Step 3

b) When $θ = - 225 °$ :

$\begin{array}{rcl} \sin (- 225 °) & = & - \sin (225 °) [S i n c e \sin (- θ) = - \sin θ] \\ = & - \sin (180 ° + 45 °) \\ = & \sin (45 °) [S i n c e \sin (180 ° + θ) = - \sin θ] \\ = & \frac{1}{\sqrt{2}} \\ \cos (- 225 °) & = & \cos (225 °) [S i n c e \cos (- θ) = \cos θ] \\ = & \cos (180 ° + 45 °) \\ = & - \cos (45 °) [S i n c e \cos (180 ° + θ) = - \cos θ] \\ = & - \frac{1}{\sqrt{2}} \\ \tan (- 225 °) & = & - \tan (225 °) [S i n c e \tan (- θ) = - \tan θ] \\ = & - \tan (180 ° + 45 °) \\ = & - \tan (45 °) [S i n c e \tan (180 ° + θ) = \tan θ] \\ = & - 1 \end{array}$

Hence

$\begin{array}{rcl} \sin (- 225 °) & = & \frac{1}{\sqrt{2}} \\ \cos (- 225 °) & = & - \frac{1}{\sqrt{2}} \\ \tan (- 225 °) & = & - 1 \end{array}$

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