Problem:
cos[arccos(-sqrt3/2)-arcsin(-1/2)]
This is what I have:
cos [(5pi/6)-(11pi/6)]
cos (-6pi/6) or -pi
giving an answer of -1
What I am wondering is should I use the (11pi/6) or should I use (-pi/6) because of the restrictions on arcsin x (-pi/2)<=x<=(pi/2)
Thank you in advance.
I would use the smaller angle. It should not matter on the final answer.
Thank you very much.
The answer doesn't change, but my trig teacher wants to see all of work, and I just wasn't sure which figure to use.
To evaluate the expression cos[arccos(-sqrt(3)/2) - arcsin(-1/2)], let's break it down step by step.
1. Let's start with the innermost expression, arccos(-sqrt(3)/2). This equation gives us the angle whose cosine is equal to -sqrt(3)/2. Since this value is negative, we know that the angle lies in either the 2nd or 3rd quadrant.
2. The value -sqrt(3)/2 corresponds to an angle of pi/6 or 11pi/6 in the unit circle. However, since we are looking for a negative value, the angle must be 11pi/6.
3. Moving to the next component, arcsin(-1/2), we want to find the angle whose sine is equal to -1/2. This value is negative, indicating that the angle lies in the 3rd or 4th quadrant.
4. The value -1/2 corresponds to an angle of -pi/6 or 7pi/6 in the unit circle. Since we are dealing with a negative value, the angle must be -pi/6.
5. Now, let's substitute the values we found into the original expression: cos[(11pi/6) - (-pi/6)]. Simplifying this expression, we get cos(12pi/6), which further reduces to cos(2pi) or cos(0).
6. Finally, cos(0) is equal to 1, so the answer to the original expression is 1.
Regarding your question about the restrictions on arcsin(x), you are correct that arcsin(x) is only defined for values of x between -1 and 1, inclusive. However, in this case, the values of -sqrt(3)/2 and -1/2 are within this range, so there are no issues with the restrictions.
Therefore, the correct answer to the expression cos[arccos(-sqrt(3)/2) - arcsin(-1/2)] is indeed 1.