+130575 The values generated by these solutions are ones associated with the pretty common "reference" angles in the first quadrant [0, pi/6, pi/4, pi/3 and pi/2]....you should commit these to memory.......and then, be able to apply them in any of the other 3 quadrants....this is a VERY important trig topic....!!!
This might help.......http://www.coolmath.com/reference/circles-trigonometry.html
+130575 I think this person wants to write an equation of a line using these points...
OK....let's get the slope, first ....where the slope, m, is defined as change in y / change in x
And from the first point to the second, y changes by -1
And from the first point to the second, x changes by - 9
So...the slope is -1 / -9 = 1/9 ....and we can use one of the points to write the equation...I'll use the first one ... so we have
y - 4 = (1/9)(x -5) simplify the right side and add 4 to both sides
y = (1/9)x - 5/9 + 4
y = (1/9)x + 31/9
Here's the graph.....https://www.desmos.com/calculator/azju8icg1f
+130575 2cosΘsinΘ=cosΘ subtract cosΘ from each side
2cosΘsinΘ - cosΘ = 0 factor
cosΘ (2sinΘ - 1) = 0
And setting each factor to 0, we have
cosΘ = 0 and this happens at pi/2 and 3pi/2
And
sinΘ = 1/2 and this happens at pi/6 and 5pi/6
It looks like you just included the first quadrant values in your solution......
Here's a graph of both sides......https://www.desmos.com/calculator/crdklgepre
