posted by George .
Find the values of x between (or equal to) 0 and 360 degrees to satisfy each equation.
I got 30 degrees as one angle, but how do I find the second one?
Is the answer 45 degrees?
Find each value:
sin(2 sin^-1 1/2)
Thanks in advance. Even a little help is appreciated.
ALWAYS draw the problem
sketch your origin and x,y axes
now you can see that for a unit hypotenuse, the opposite side (y) is positive in guadrants one and two,\.
so 30 degrees is one answer
and 30 degrees above the -y axis is the other place where sin = 1/2
that is of course 180 - 30 = 150 degrees
now if cotangent is one, tangent is one.
so sure, 45 degrees will do.
BUT where else is y/x positive?
It is where both x and y are negative, quad 3
so 180 + 45 = 2556 is second answer
I think you can take it from there. Sketch your four quadrants and then your x and y
sines are positive in the first and second quadrant. If sin A = 1/2, A can be wither 30 or 150 degrees. The absolute value of sin, cost and tan etc. are determined by the angle with the horizontal axis.
cot and tan are negative in the second and fourth quadrants, and positive in the first and third quadrants.
cot^-1(1) = 45 or 225 degrees
cosine is negative in second and third quadrants.
cos-1(-1/2) = 120 and 240 degrees (60 degrees from the -x axis)
sin(2 sin^-1 1/2) = sin 2*30 or sin 2*150 = +0.866 or -0.866, since there are twqo possible values for sin^-1 (1/2)
I got 30 degrees as one angle, but how do I find the second one?"
There is only one angle and it is 30 degrees. Don't confuse arcsin(1/2) with the set of the solutions of the equation sin(x) = 1/2.
It is similar to solving the equation:
x^2 = 4
and the squareroot function, in this case sqrt(4). The solution of the equation x^4 = 4 is not unique, there are two solutions: x = 2 and x = -2. So, there are two possible inverse functions that one can define. One has to make some choice. The squareroot function is defined as the postive solution of the equation. So sqrt(4) = 2 and not -2.
You can imagine what a terrible mess it would be if the two possible definitions were both used.
Similarly, one has defined the arcsin function such that it gives ONE of the solutions of the equation
sin(x) = y. By definition arcsin(y) is that solution of sin(x) = y that lies in the range between -pi/2 and pi/2.
but the question asked:
" Find the values of x between (or equal to) 0 and 360 degrees to satisfy each equation.
Which is not exactly what is arcsin(1/2)