How do you find the exact value of sin^-1 0? - Socratic
Sep 8, 2016 · sin^-1(0)=npi where n is an integer. Let's start with the unit circle: The sine ratio is the opposite divided by the hypotenuse: sin="opp"/"hyp" sin^-1 asks us what the angle is that produces a ratio of a certain value. In this case, that ratio equals 0. For the ratio to be 0, the opposite side must equal 0 (if the hypotenuse were 0, the fraction would be undefined). What …
How do you evaluate #sin^-1(0)#? - Socratic
Jan 3, 2017 · 0 sin^-1(0) = x Take sin to the left and to the right. sin sin^-1 (0) = sin x = 0 x = 0 + k pi, because sin k pi = 0 But what is the domain of f(x) = sin^-1 x ?
How do you find the exact value of #sin^-1(-0.5)#? - Socratic
Apr 15, 2015 · Yes, t = π 6. But we want sint = − 1 2, so we want an answer in quadrant 3 or 4. sin−1(x) always gives us a t between − π 2 and π 2, so we want t = − π 6. sin−1(− 0.5) = π 6. When asked to evaluate inverse trigonometric functions, first ask yourself, "Is the answer a …
How do you calculate #sin^-1(0.1)#? - Socratic
Apr 28, 2018 · Find arcsin (0.1) by calculator. sin x = 0.1 --> arc # x1 = 5^@74# The unit circle gives another arc x2 that has the same sin value (0.1):
How do you find the exact value of #sin^-1(-1)#? - Socratic
Sep 10, 2016 · The range of arcsin is defined as between − π 2 radians and π 2 radians. arcsin (x) is the one angle within this range whose sine is x. The one angle between − π 2 and π 2 with a sin(x) = − 1 is x = − π 2. On the unit circle, a sin of -1 falls on the negative x …
How do you simplify #sin^-1(0.5)#? - Socratic
Apr 15, 2016 · Explanation: Sin is positive in Quadrants 1 and 2 and all its coterminals. sin−1(0.5) = π 6 +2πn, 5π 6 +2πn. Answer link. iOS.
What is #tan(sin^-1(0))#? - Socratic
Aug 29, 2015 · How do you use inverse trigonometric functions to find the solutions of the equation that are in
What is Cos(Sin^-1(0)) - Socratic
Mar 10, 2018 · 1 Hint: color(red)(sin^(-1)(sint)=t,where,t in[-1,1]) x=cos(sin^(-1)(0)), but, color(red)(sin0=0) x=cos(color(red)(sin^(-1)(sin0))),where, 0in[-1,1] :.x=cos(color(red ...
How do you calculate #sin^-1 (-0.4)#? - Socratic
Mar 5, 2018 · 1 Answer. sin x = - 0.4. Calculator gives -->. x = 203^@58 x = - 23^@58, or x = 336^@42 (co-terminal) sin^-1 (-0.4) --> arc x whose sin is (-0.4) sin x = - 0.4. Calculator gives --> x = - 23^@58, or x = 336^@42 (co-terminal) Unit circle gives another x that has the same sin value (- 0.4) x = 180 - (- 23.58) = 180 + 23.58 = 203^@58.
What is the limit of sin (1/x) as x approaches 0? - Socratic
Jun 22, 2017 · The limit does not exist. To understand why we can't find this limit, consider the following: We can make a new variable h so that h = 1/x. As x -> 0, h -> oo, since 1/0 is undefined. So, we can say that: lim_(x->0)sin(1/x) = lim_(h->oo)sin(h) As h gets bigger, sin(h) keeps fluctuating between -1 and 1. It never tends towards anything, or stops fluctuating at …