Trigonometry Question
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Grade 11 Trigonometry CCSS: HSF-TF.A.4
Which of the following reasons allows one to conclude that sin(Θ) is odd?
- It has been shown that sin(-ω)=-sin(ω) for 0≤ω≤2π. Since sine is periodic with a period of 2π, this result must be true for ω∈ℝ.
- It has been shown that sin(-ω)=-sin(ω) for 0≤ω≤2π. Since -1≤sin(ω)≤1 is true not only for values between 0 and 2π, but for all real values of ω, it follows that sin(-ω)=-sin(ω) for ω∈ℝ.
- Since the values of sine from the first quadrant are simply repeated in the other quadrants (except for being sometimes negative), all that one needed to confirm was that sin(-ω)=-sin(ω) for 0≤ω≤π2, and then it must be true for ω∈ℝ.
- Using circle A, it has been seen that, for 0≤ω≤2π, the sine of ω is always greater than or equal to zero, while the sine of -ω is always less than or equal to zero. Therefore, since sine is periodic with period 2π, this result must be true for ω∈ℝ.