Verify each identity.
a) arcsin(-x) = -arcsin x;
abs. val. (x) is less than or
equal to 1.
b) arccos(-x) = pi - arccos x;
abs. val. (x) is less than or
equal to 1.
To verify each identity, we will use the properties of inverse trigonometric functions. Here's how you can prove each identity:
a) To prove arcsin(-x) = -arcsin(x), we need to show that the left-hand side is equal to the right-hand side.
Starting with the left-hand side: arcsin(-x)
1. Recall that the arcsin function returns an angle whose sine is equal to its input.
2. Since we have -x as the input, we want to find an angle whose sine is -x.
3. We know that the sine function is an odd function, meaning sin(-x) = -sin(x). Therefore, we can rewrite the left-hand side as -sin(x).
The right-hand side: -arcsin(x)
1. The minus sign in front of arcsin(x) can be moved inside as -sin(x), using the same property mentioned above.
Hence, both sides are equal, and the identity arcsin(-x) = -arcsin(x) holds true for absolute value (|x|) less than or equal to 1.
b) To prove arccos(-x) = pi - arccos(x), we need to show that the left-hand side is equal to the right-hand side.
Starting with the left-hand side: arccos(-x)
1. The arccos function returns an angle whose cosine is equal to its input.
2. Since we have -x as the input, we want to find an angle whose cosine is -x.
3. We know that the cosine function is an even function, meaning cos(-x) = cos(x). Therefore, we can rewrite the left-hand side as arccos(x).
The right-hand side: pi - arccos(x)
1. The subtraction between pi and arccos(x) is possible because arccos(x) returns an angle in the range [0, pi].
2. By replacing arccos(x) with arccos(x) using the even property, we get pi - arccos(x).
Hence, both sides are equal, and the identity arccos(-x) = pi - arccos(x) holds true for absolute value (|x|) less than or equal to 1.