Find the exact value of:
sin [2sin^-1 (-sq rt (3)/2)]
how do i answer this question? i don't know how to find the answer.
start with sin^-1 (-√3/2)
which is 240 degrees or 300 degrees
so 2sin^-1 (-√3/2) is 480 or 600 degrees
then the sine of that ...
sin 480 = sin 120 = sin 60 = √3/2
sin 600 = sin 240 = - sin 60 = -√3/2
To find the exact value of sin [2sin^-1 (-√3/2)], we can break it down into smaller steps. Let's start by understanding what sin^-1 (-√3/2) means.
The notation sin^-1 (x) represents the inverse sine function, also known as arcsine. It gives us the angle whose sine is equal to x. In this case, sin^-1 (-√3/2) gives us the angle θ, where sin(θ) = -√3/2.
By drawing a right triangle with an angle θ and the opposite side of length √3 and the hypotenuse of length 2, we can find θ to be -π/3 (or -60 degrees).
Now, let's move on to sin [2sin^-1 (-√3/2)]. The value inside the brackets, sin^-1 (-√3/2), is -π/3. We need to find sin (2 * (-π/3)).
Using the double-angle identity for sine, sin(2θ) = 2sin(θ)cos(θ), we can calculate sin (2 * (-π/3)) as follows:
sin (2 * (-π/3)) = 2sin(-π/3)cos(-π/3)
Since sin(-π/3) = -√3/2 and cos(-π/3) = 1/2, we can substitute the values:
sin (2 * (-π/3)) = 2 * (-√3/2) * (1/2) = -√3/2
Therefore, the exact value of sin [2sin^-1 (-√3/2)] is -√3/2.