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I evaluated csc(-630 °) = csc(90°) and understand it is a quadrantal, but I do not understand why it is 1 nor why it is +1 and not -1. Can someone explain?

user137452
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    $\csc 90^{\circ} = \frac{1}{\sin 90^{\circ}} = \frac{1}{1} = 1$. So, is your question why is $\sin 90^{\circ} = 1$? – D_S Jul 13 '14 at 01:10
  • When an angle is between $0$ and $90$ degrees, I assume you understand how $\sin, \cos, \tan$ etc. of that angle are defined. To evaluate trigonometric functions at arbitrary angles, you need to look at the unit circle. Are you confused about how trig functions are defined for arbitrary angles? – D_S Jul 13 '14 at 01:13
  • The reference angle is at 90 degrees. Thus, is not in a quadrant. My questions is why is it 1 and why is it positive? – user137452 Jul 13 '14 at 01:16

2 Answers2

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Hint: We have $-630=-2(360)+90$. So every trigonometric function is the same at $-630$ degrees as at $90$ degrees.

André Nicolas
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Given an angle $\theta$, you have a corresponding ray from the origin going to a point $(x,y)$ on the unit circle. The angle $\theta$ is obtained by going counterclockwise from the point $(1,0)$. For such an angle $\theta$ and a corresponding point $(x,y)$ on the unit circle, $\sin \theta$ is defined to be $y/1 = y$.

You can see this agrees with the definition of $\sin$ for right triangles when $\theta$ is between $0$ and $90$ degrees. In any case $y$ is obviously positive when $\theta$ is strictly between $0$ and $180^{\circ}$, so $\sin y$ will be positive. And when $\theta = 90^{\circ}$, your line from the origin shoots straight up and meets the circle at $(0,1)$. So, $\sin 90^{\circ} = 1$.

D_S
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  • Ok. If SINE is the y value and COSINE is the x value, why is it that sin(180°) is -1 and not 0? – user137452 Jul 13 '14 at 01:42
  • $\sin 180^{\circ}$ is $0$ :)

    On the other hand, going around $270^{\circ}$ gets you to the point $(0,-1)$, so $\sin 270^{\circ} = -1$.

     – D_S Jul 13 '14 at 01:44
  • Not according to my textbook. – user137452 Jul 13 '14 at 01:48
  • Please evaluate sin(3pi/2) = – user137452 Jul 13 '14 at 01:50
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    Your textbook is wrong. I'm sorry it seems to be a source of confusion for you, but I promise you that $\sin 180^{\circ} = 0$. On the other hand, $\cos 180^{\circ}$ IS $-1$. – D_S Jul 13 '14 at 01:50
  • $2 \pi$ radians is the same as $360$ degrees. Since $3 \pi/2 = \frac{3}{4} \cdot (2 \pi)$, we have that $\frac{3 \pi}{2}$ radians is the same as $\frac{3}{4} \cdot 360 = 270$ degrees. So $\sin$ of $3 \pi /2$ radians is equal to $\sin 270^{\circ} = -1$. – D_S Jul 13 '14 at 01:54
  • You are correct it is 270 thus -1. I converted the radians to degrees and made mistake. Thanks. – user137452 Jul 13 '14 at 01:57
  • You're welcome! If you are satisfied with my or someone else's answer, please "accept" it by clicking the check mark so your post will no longer appear on the list of unanswered questions. – D_S Jul 13 '14 at 02:00