4/4/2023 0 Comments Iunit circle chart![]() ![]() The graph of the unit circle with radius = 1 and points of intersection with X and Y axes In other words, the center is put on a graph where the X and Y axes cross. The unit circle is a circle with a radius of one unit with its center placed at the origin. What is the Unit Circle and how is it used? Let's learn more about it and test our understanding with a handy trigonometric calculator I created at the end of the article. It removes the need for memorizing different values and allows the user to simply derive different results for different cases. The key to its usefulness is its simplicity. The trigonometric ratio defining the tangent (opposite over adjacent) may be less intuitive from the unit circle, but I hope you can see from the unit circle why the tangent of an angle is equal to the sine of the angle over the cosine.The unit circle is a useful visualization tool for learning about trigonometric functions. Similarly, the sine is the length of the opposite side (which is the length of the y-coordinate) divided by the length of the hypotenuse (which is 1). ![]() The cosine is the length of the adjacent side (which is the x-coordinate) divided by the length of the hypotenuse (which is 1). This means you can figure out the sine and cosine of the angle that the hypotenuse and the x-axis make: So you have a right triangle where one side length is the x-coordinate and one side length is the y-coordinate, and the hypotenuse is equal to 1. Take a little time drawing right triangles were the hypotenuse connects the middle of the unit circle to a point on the circle and see if you can convince yourself that this is the case. Because that's the radius of the circle! And the length of the other two sides are just going to be the x and y coordinates of the point on the unit circle. The cool thing about a unit circle (a circle with radius one), is that if you draw a right triangle where the hypotenuse of the triangle connects a point on the unit circle with the point (0, 0), the length of the hypotenuse is always going to be one. Here I think would be my answer to this question, if something is wrong, please let me know and we can figure out that~ Under this situation, we know as long as m (the angle theta) satisfies sin2m=-1, it is the solution and now 2m= 3*pi/2 + 2*k*pi, k is an integer so m=3*pi/4 + k*pi, k is an integer we now see that the terminal side of m is the bisector of the 2nd (II) and 4th (IV) quadrant, if in the 2nd quadrant, sinm=sqrt(2)/2 and cosm= - sqrt(2)/2 we have sinm-cosm=sqrt(2) if in the 4th quadrant, sinm= - sqrt(2)/2 and cosm= sqrt(2)/2, we have sinm-cosm= - sqrt(2). If x doesn't equal to 0, we divide x on both sides and we have: figure out the only situation to hold the equation true is that -1/sinm*cosm=2, that means (-1*2)/(sin2m)=2 and sin2m must be -1 to keep this true. Now it turns into a problem solving x: if x=0, it satisfies the equation and this means sinm=cosm these angles can be the bisector of the 1st (I) and 3rd(III) quadrant they can be pi/4 and 3pi/4 as you mentioned before and 2k*pi applied on them as well, under this condition, we choose 0 as the answer Let x = sinm - cosm, which is the value we try to find we have -x/(sinm*cosm)=2x Then move the terms and change it into: 1/sinm -1/cosm = 2(sinm - cosm) simplify this as: (cosm-sinm)/(sinm*cosm)=2(sinm-cosm) Then divide sinm on both sides (if sinm=0, the equation cannot be satisfied since left=0, right=0-1=-1, not equal to the left), then we have: 2cosm=1/cosm -1/sinm +2sinm Actually not just these two radians but a set of them are the solutions to the 1st equation Check this:įirst, I use m to represent theta, sqrt(2) as the square root of 2, for simplification: so we know sin(2m)=tan(m)-cos(2m) from the given facts, then we use the identities, we have left=sin(2m)=2*sinm*cosm=right=(sinm/cosm) - , If you try theta= pi/4 or negative pi/4 the first equation holds in both situations and you get choice A and C respectively. Thank you for this question, this is a great one I think. ![]()
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