Inverse trigonomtric functions   arcsin 3t > pi/6

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to substitute `arcsin (0.5)` for `pi/6` , to the right side, such that:

`arcsin (3t) > arcsin (0.5) `

You need to remember that the monotony of the arcsine function tells that it increases over R, using the definition of an increasing function yields:

`x_1 < x_2 => f(x_1) < f(x_2)`

Reasoning by analogy, yields:

`arcsin (0.5) < arcsin (3t) => 0.5 < 3t => (0.5)/3 < t`

Hence, evaluating the given inequality yields that `t in ((0.5)/3,oo).`

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

From the information you've given, I suppose that you want to determine the interval of values of t, for the inequality to hold.

We consider the inverse function arcsin 3t > pi/6.

We'll apply sine functin both sides:

sin (arcsin 3t) > sin pi/6

Since the sine function is increasing, we'll keep the direction of the inequality:

3t > 1/2

We'll divide by 3 both sides:

t> 1/6

The interval of admissible values of t, for the inequality to hold, is: (1/6 ; +infinite).

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