![]() ![]() Since tan is positive in both quadrants 1 and 3, the sign on some of the trig functions might differ. Finally, use sin 2(alpha) + cos 2(alpha)=1 to get sin(alpha). ![]() Use sec(alpha)= sqrt(tan 2(alpha)+1) and then use that to get cos(alpha) since cos(alpha) = 1/sec(alpha). To summarize, use cot(alpha) = 1/tan(alpha) to get cot(alpha). The triangle shaded blue illustrates the identity +, and the red triangle shows that +. The reciprocal identities arise as ratios of sides in the triangles where this unit line is no longer the hypotenuse. Finally plug that back into sin 2(alpha) + cos 2(alpha)=1 to solve for cos(alpha). For sin, cos and tan the unit-length radius forms the hypotenuse of the triangle that defines them. Since sec(alpha) = 1/cos(alpha), it = the reciprocal of sqrt(37)/6 which is 6*sqrt(37)/37. It might be negative depending on the original angle but let's just assume it is positive for now. ![]() The problem with plugging the 1/6 into that is you do not know which part of the circle or graph it is in so sec(alpha) upon solving would be sqrt = sqrt = sqrt(37)/6. The formula sin 2(alpha) + cos 2(alpha)=1 can be divided through by cos 2 to give tan 2(alpha) + 1= sec 2(alpha). First, cot(alpha) = 1/tan(alpha) so 1 over 1/6 is 6. If the tangent of an angle is 1/6, there are a few ways to find the other trig values. ![]()
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