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**Unformatted text preview: **e α, we subtract: α = 2π − θ = 2π − 11π = π .
6
6
Since θ is a Quadrant IV angle, cos(θ) > 0 and sin(θ) < 0, so the Reference Angle Theorem
√
gives: cos 11π = cos π = 23 and sin 11π = − sin π = − 1 .
6
6
6
6
2
y y 1 1 θ = 225◦ π
6 x 1 45◦ θ= Finding cos (225◦ ) and sin (225◦ ) 1 x 11π
6 11π
6 Finding cos and sin 11π
6 π
π
3. To plot θ = − 54 , we rotate clockwise an angle of 54 from the positive x-axis. The terminal
π
side of θ, therefore, lies in Quadrant II making an angle of α = 54 − π = π radians with
4
respect to the negative x-axis. Since θ is a Quadrant II angle, the Reference Angle Theorem
√
√
π
π
gives: cos − 54 = − cos π = − 22 and sin − 54 = sin π = 22 .
4
4
π
π
4. Since the angle θ = 73 measures more than 2π = 63 , we ﬁnd the terminal side of θ by rotating
π
one full revolution followed by an additional α = 73 − 2π = π radians. Since θ and α are
3 coterminal, cos 7π
3 = cos π
3 = 1
2 and sin 7π
3 = sin π
3 √ = 3
2. y y 1 1 θ= π
4 1 π
3...

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