simplify sin(2 arccos(x))
We can simplify sin(2 arccos(x)) using the double angle trigonometric identity for sine:
sin(2θ) = 2sin(θ)cos(θ)
First, let's find sin(arccos(x)). We know that arccos(x) gives us the angle whose cosine is x. So, if we have a right triangle with an angle θ whose cosine is x, then the opposite side would have a length of √(1 - x^2), and the hypotenuse would have a length of 1. Therefore, sin(arccos(x)) = √(1 - x^2)/1 = √(1 - x^2).
Now, using the double angle identity, we have:
sin(2 arccos(x)) = 2sin(arccos(x))cos(arccos(x))
= 2(√(1 - x^2))(x)
= 2x√(1 - x^2)
So, sin(2 arccos(x)) simplifies to 2x√(1 - x^2).
To simplify the expression sin(2arccos(x)), we can use the double angle formula for sin(2θ).
The double angle formula for sin(2θ) is:
sin(2θ) = 2sin(θ)cos(θ)
In this case, we have arccos(x) as θ. Therefore, θ = arccos(x).
Substituting θ into the double angle formula, we get:
sin(2(arccos(x))) = 2sin(arccos(x))cos(arccos(x))
Using the trigonometric identity sin(arccos(x)) = √(1 - x^2) and cos(arccos(x)) = x, we can simplify further:
sin(2(arccos(x))) = 2sin(arccos(x))cos(arccos(x))
= 2(√(1 - x^2))x
= 2x√(1 - x^2)
So, the simplified expression for sin(2arccos(x)) is 2x√(1 - x^2).
To simplify the expression sin(2 arccos(x)), we can use the double angle formula for sine.
The double angle formula for sine states that sin(2θ) = 2 sin(θ) cos(θ).
In this case, our angle is 2 arccos(x), so let's substitute θ with arccos(x):
sin(2 arccos(x)) = 2 sin(arccos(x)) cos(arccos(x))
Now, we can simplify further.
sin(arccos(x)) represents the sine of an angle whose cosine is x. By drawing a right triangle, we can determine that the value of the sine in this case is equal to the square root of (1 - x^2).
cos(arccos(x)) represents the cosine of an angle whose cosine is x. This is simply x.
Substituting these values into our equation, we get:
sin(2 arccos(x)) = 2 √(1 - x^2) * x
So, the simplified expression for sin(2 arccos(x)) is 2x√(1 - x^2).