Trigonometry Question

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Let circle A be the unit circle centered at the origin. Assume that

$x_1, x_2, x_3, x_4, y_1, y_2, y_3, y_4$ are all real numbers that are greater than or equal to zero.

Grade 11 Trigonometry CCSS: HSF-TF.A.4

Looking at

$pi < beta <= (3pi)/2$ and

$(3pi)/2 < gamma <= 2pi$ , where both angles are in standard position, and their terminal arms intersect circle A at

$(-x_3, -y_3)$ and

$(x_4, -y_4)$ respectively, which of the following explains why the sine of an angle equals the negative sine of the negative of that angle for angles between

$pi$ and

$2pi ?$

Since sine is periodic, whatever is true for the sine function for values between $0$ and $pi$ must also be true for values between $pi$ and $2pi$ .
Using the same reasoning as before, for angle $beta$ it follows that $sin(beta) = -y_3 = -sin(-beta)$ and for angle $gamma$ that $sin(gamma) = -y_4 = -sin(-gamma)$ .
As seen before, the sine of an angle in one quadrant is always the opposite of the sine of an angle in an adjacent quadrant. Thus $sin(beta) = -sin(-beta)$ and $sin(gamma) = - sin(-gamma)$ .
Looking at angles between $pi$ and $2pi$ , the sine of these angles is always negative. Therefore, $sin(beta) = - sin(-beta)$ and $sin(gamma) = - sin(-gamma)$ .

Question Info

Trigonometry Question

Grade 11 Trigonometry CCSS: HSF-TF.A.4