Question 624826
The key to both of these problems is knowing the range of the various inverse trig functions.<br>
For the first problem, the inverse tan function has a range of 0 to {{{pi}}}. And since we have been given a positive tan value, 7/24, the angle must be in the range 0 to {{{pi/2}}}, i.e. a first quadrant angle. Since sin is also positive in the first quadrant we will end up with a positive result.<br>
The only question remaining is: What positive number. Since we're asked for exact values we must not use the Trig buttons on our calculators. To find our solution, we imagine (or draw) a right triangle and pick one of the acute angles. Since tan is opposite over adjacent, make the opposite side 7 and the adjacent side 24. Since sin is opposite over hypotenuse, we need to find the hypotenuse. Use the Pythagorean Theorem to find the hypotenuse:
{{{c^2 = 7^2 + 24^2}}}
You should find the hypotenuse to be 25. So {{{sin(tan^(-1)(7/24)) = 7/25}}}<br>
For the second problem one might think that the inverse cos of the cos of {{{4pi/3}}} would be {{{4pi/3}}}! But this would not be correct because the range of the inverse cos is 0 to {{{pi}}} and {{{4pi/3}}} is not in this range. So whatever answer we get, it must be between 0 and {{{pi}}}. {{{4pi/3}}} is an angle which terminates in the 3rd quadrant with a reference angle of {{{4pi/3 - pi = pi/3}}}. Since {{{cos(pi/3) = 1/2}}} and since cos is negative in the 3rd quadrant, {{{cos(4pi/3) = -1/2}}}<br>
Substituting this into our expression we get:
{{{cos^(-1)(-1/2)}}}
Now we just have to figure out what angle, between 0 and {{{pi}}} has a cos of -1/2?. Well the reference angle is still {{{pi/3}}}. And in the range 0 to {{{pi}}} only 2nd quadrant angles have negative cos values. So the angle we are looking for is the second quadrant angle with a reference angle of {{{pi/3}}}:
{{{pi - pi/3 = 2pi/3}}}
So {{{cos^(-1)(cos(4pi/3)) = 2pi/3}}}