Explain why sin^-1[sin(3pi/4)] does not = 3pi/4 when y=sin(x) and y=sin^-1(x) are inverses.
Any help on this question is greatly appreciated. Thank you!
sin(3PI/4) is a specific number. The inverse of that number can track to several angles, all of which have the same value of sine.
Example; arcsin(sin30deg).
well sin 30 deg=.5
but arcsin(.5) can be 30 deg, 150 deg, and so on...
well sin 3 pi/4 = +1/sqrt 2 = +sqrt 2/ 2 in quadrant 2
BUT
sin pi/4 = +sqrt2 / 2 in quadrant 1
sin(3pi/4) = sin(pi/4)
sin^-1 (sqrt2 /2) = 45 degrees and sin^-1(sqrt2/2) = 135 degrees
so for example
sin^-1 (sin3pi/4) = pi/4
To understand why sin^-1[sin(3pi/4)] does not equal 3pi/4 when y=sin(x) and y=sin^-1(x) are inverses, we need to understand how inverse trigonometric functions work.
Inverse trigonometric functions such as sin^-1(x) or arcsin(x) represent the angle whose sine value is equal to x. In other words, sin^-1(x) gives the angle whose sine is x.
However, it's important to note that inverse trigonometric functions have a restricted domain. The domain of sin^-1(x) is [-1, 1]. Therefore, the output of sin^-1(x) will always be an angle between -pi/2 and pi/2.
Now, let's apply this to the expression sin^-1[sin(3pi/4)]. We have sin(3pi/4), which is equal to sqrt(2)/2. Therefore, we are finding the angle whose sine is sqrt(2)/2.
Since the sin^-1(x) function only returns values between -pi/2 and pi/2, the output of sin^-1[sin(3pi/4)] will be an angle within that range.
sin(3pi/4) equals sqrt(2)/2, which is in the restricted domain of sin^-1(x). Therefore, sin^-1[sin(3pi/4)] will give us an angle within the range of -pi/2 to pi/2.
To determine the exact value, we can find the angle whose sine is sqrt(2)/2. This will give us the principal value, which gives us an angle within the restricted range.
In this case, the angle whose sine is sqrt(2)/2 is pi/4 (45 degrees). Therefore, sin^-1[sin(3pi/4)] equals pi/4, not 3pi/4.
In summary, sin^-1[sin(3pi/4)] does not equal 3pi/4 because sin^-1(x) only returns values between -pi/2 and pi/2, and the angle whose sine is sqrt(2)/2 is pi/4, not 3pi/4.
Sure! To understand why sin^-1[sin(3π/4)] does not equal 3π/4, we need to understand the concept of inverse trigonometric functions and their domains.
In trigonometry, the sine function (sin(x)) takes an angle x as input and returns the corresponding y-value on the unit circle. The domain of the sine function is all real numbers, while the range is [-1, 1].
On the other hand, the inverse sine function (sin^-1(x) or arcsin(x)) takes a y-value as input and returns the corresponding angle in radians. The domain of the inverse sine function is the range of the sine function, which is [-1, 1], and the range is [-π/2, π/2].
Now, let's analyze sin^-1[sin(3π/4)]:
1. Start with the inner function sin(3π/4). Since 3π/4 is an angle in the second quadrant, the sine of this angle will be positive (√2/2).
2. Next, apply the inverse sine function to the result of step 1. sin^-1(sin(3π/4)) = sin^-1(√2/2) = ?
Here's where the problem arises. The inverse sine function can only return angles in the range [-π/2, π/2]. In this range, there are two angles whose sine is √2/2, namely π/4 and 3π/4.
So, sin^-1(√2/2) is ambiguous and can be either π/4 or 3π/4. Both angles have the same sine value (√2/2), but we need to consider the range of the inverse sine function.
Therefore, sin^-1[sin(3π/4)] does not equal 3π/4 because the inverse sine function can only return values in the range [-π/2, π/2].
If we want to evaluate sin^-1[sin(3π/4)], we have to write it as sin^-1[sin(3π/4)] = π/4, the angle within the defined range of the inverse sine function that gives the same sine value.