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kittiwitti1
 one year ago
As θ increases from 0° to 90°, the value of cos θ tends toward which of the following?
Answers available: 0 and 1. (Putting pos/neg infinity is wrong so I just need to decide which).
kittiwitti1
 one year ago
As θ increases from 0° to 90°, the value of cos θ tends toward which of the following? Answers available: 0 and 1. (Putting pos/neg infinity is wrong so I just need to decide which).

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mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1441598341152:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3The cosine of an angle is the xcoordinate of the point of intersection of the terminal side of the angle and the unit circle.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1441598453980:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3When theta = 0 deg, cos(theta) = 1, since the xcoordinate of the point of intersection of the terminal side of of a zerodegree angle (in standard position) and the unit circle is 1.

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0I mean, the xvalues are steadily decreasing so I assumed 0, but I am just making sure.

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3dw:1441598614600:dw

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3As theta goes from 0 deg to 90 deg, the xcoordinate goes from 1 to 0.

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0@mathstudent55 could you help me with this one as well? cos θ = 1/2 and θ terminates in QIV. So far I have this: http://prntscr.com/8dhn18

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\sin(\theta)=\pm\sqrt{1\frac{1}{4}}=\pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}\] Since theta is in 4th quadrant, sin will be negative \[\sin(\theta)=\frac{\sqrt{3}}{4}\] There is a reason why you are given in which quadrant theta belongs to, so you can apply proper signs!

kittiwitti1
 one year ago
Best ResponseYou've already chosen the best response.0Thanks @Nishant_Garg :]

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.3@Nishant_Garg \(\large \sin(\theta)=\pm\sqrt{1\frac{1}{4}}=\pm\sqrt{\frac{3}{4}}=\pm\frac{\sqrt{3}}{2}\) \(\large \sin(\theta)=\frac{\sqrt{3}}{\huge \color{red}{4}}\) ?
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