Sin and cos!

Joined
Jul 31, 2012
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sin <A = sin <B where <B and <A do not equal one another. <A= 60 degrees.

I do not understand how obtain <B, and what would the first couple steps be? Also, would it be easier to represent this as a radian or in degree? Thanks you very much!

By the way: Could you suggest any more of these kinds of problems so I could practice? Thank you so much and for your time!
 
How to answer this depends upon exactly how you are defining sine and cosine. The definitions in terms of ratios of a right triangle aren't sufficient because we want them defined for all real numbers. The simplest way to do that is: Draw a unit circle on an xy-coordinate system. Starting at (1, 0), measure counterclockwise around the circumference of the circle a distance t (equivalently, measure t radians at the center of the circle). The ending point has coordinates (cos t, sin t). So, to find another value of t such that sin t= sin pi/3 (pi/3 radians is 60 degrees), mark 60 degrees on that circle and draw the horizontal line through that point. The horizontal line has constant "y" value so the second point where the line crosses the circle is the other t that gives the same sine.

You should be able to see that the the second angle is 180- 60 degrees.
 
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I suggest the following problem if you still need extra practice: Draw the unit circle from memory and plot the 16 points of interest(multiples of 30 and 45 degrees) including their coordinates.

If you can do that and understand the points are of form (cos t, sin t) these types of problems should become easy for you.
 
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