im really stuck on these, can anyone help me and also explain how to solve them:
Find the exact values of s in the given intervals that have the given circular function values.
1. [pi/2, pi]; sin s=sqrt2/2
2. [pi/2, pi]; cos s=-sqrt3/2
thnks:)
sqrt 2 is the legs of a 45,45, 90 triangle, hypotenuse is 2
sqrt 3 is the side opposite the 60 degree angle in a 30,60, 90 triangle. (legs are 1 and sqrt 3, hypotenuse is 2)
Learn those right triangles by heart. They come up constantly.
how does that help me find s though..
Sure, I'd be happy to help you with these problems and explain how to solve them.
1. In the given interval \([ \frac{\pi}{2}, \pi ]\) and the circular function value sin s = \(\frac{\sqrt{2}}{2}\). To find the exact values of s, we need to look for angles that satisfy this condition.
First, let's recall the special angles on the unit circle. The angle \(\frac{\pi}{4}\) corresponds to the point \((\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2})\) on the unit circle. Since sin s = \(\frac{\sqrt{2}}{2}\), we can see that s = \(\frac{\pi}{4}\) satisfies the equation sin s = \(\frac{\sqrt{2}}{2}\).
However, \(\frac{\pi}{4}\) is not in the given interval [ \(\frac{\pi}{2}\), \(\pi\) ]. Since sin function is negative in the interval [ \(\frac{\pi}{2}\), \(\pi\) ], we need to find the reference angle in the first quadrant and calculate the corresponding angle in the second quadrant.
The reference angle for sin s = \(\frac{\sqrt{2}}{2}\) is \(\frac{\pi}{4}\). To find the angle s in the second quadrant, we can subtract the reference angle from \(\pi\).
s = \(\pi - \frac{\pi}{4}\) = \(\frac{3\pi}{4}\)
Therefore, in the interval [ \(\frac{\pi}{2}\), \(\pi\) ], the exact value of s that satisfies sin s = \(\frac{\sqrt{2}}{2}\) is \(\frac{3\pi}{4}\).
2. In the given interval [ \(\frac{\pi}{2}\), \(\pi\) ] and the circular function value cos s = -\(\frac{\sqrt{3}}{2}\). To find the exact values of s, we need to look for angles that satisfy this condition.
The angle \(\frac{\pi}{6}\) corresponds to the point \((\frac{\sqrt{3}}{2}, \frac{1}{2})\) on the unit circle. Since cos s = -\(\frac{\sqrt{3}}{2}\), we can see that s = \(\frac{\pi}{6}\) satisfies the equation cos s = -\(\frac{\sqrt{3}}{2}\).
However, \(\frac{\pi}{6}\) is not in the given interval [ \(\frac{\pi}{2}\), \(\pi\) ]. Since cos function is negative in the interval [ \(\frac{\pi}{2}\), \(\pi\) ], we need to find the reference angle in the second quadrant and calculate the corresponding angle in the third quadrant.
The reference angle for cos s = -\(\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\). To find the angle s in the third quadrant, we can subtract the reference angle from \(\pi\).
s = \(\pi - \frac{\pi}{6}\) = \(\frac{5\pi}{6}\)
Therefore, in the given interval [ \(\frac{\pi}{2}\), \(\pi\) ], the exact value of s that satisfies cos s = -\(\frac{\sqrt{3}}{2}\) is \(\frac{5\pi}{6}\).
I hope this explanation helps you understand how to solve similar problems in the future. Let me know if you have any further questions!