simplify the following expression sin(v+x)-sin(v-x)?
Rewrite the expression as an algebraic expression in x. sin(2cos^-1x)
sin(v+x) = sinv*cosx + cosv*sinx
sin(v-x) = sinv*cosx - cosv*sinx
subtract to get
2cosv*sinx
sin(cos^-1 x) = √(1-x^2)
by definition, cos(cos^-1 x) = x
sin(2cos^-1 x) = 2sin(cos^-1 x)cos(cos^-1 x)
= 2√(1-x^2) * x
= 2x√(1-x^2)
To simplify the expression sin(v+x) - sin(v-x), you can use the identity for the difference of two sines:
sin(A) - sin(B) = 2*cos((A + B)/2)*sin((A - B)/2)
In this case, you have A = v + x and B = v - x. Plugging these values into the identity, you get:
sin(v + x) - sin(v - x) = 2*cos(((v + x) + (v - x))/2)*sin(((v + x) - (v - x))/2)
Simplifying further:
sin(v + x) - sin(v - x) = 2*cos((2v)/2)*sin(2x/2)
= 2*cos(v)*sin(x)
So the simplified expression is 2*cos(v)*sin(x).
Now, let's rewrite the expression sin(2*cos^-1(x)) as an algebraic expression in x.
The expression sin(2*cos^-1(x)) can be rewritten using the double angle formula for sine:
sin(2θ) = 2*sin(θ)*cos(θ)
In this case, θ = cos^-1(x). Plugging this value into the double angle formula, you get:
sin(2*cos^-1(x)) = 2*sin(cos^-1(x))*cos(cos^-1(x))
Now, using the identity sin(θ) = √(1 - cos^2(θ)), we can rewrite sin(cos^-1(x)) as:
sin(cos^-1(x)) = √(1 - cos^2(cos^-1(x)))
= √(1 - x^2)
Similarly, cos(cos^-1(x)) is simply x.
So the expression sin(2*cos^-1(x)) can be simplified to:
sin(2*cos^-1(x)) = 2*sin(cos^-1(x))*cos(cos^-1(x))
= 2*√(1 - x^2)*x
Therefore, the algebraic expression for sin(2*cos^-1(x)) is 2√(1 - x^2)*x.
To simplify the expression sin(v+x) - sin(v-x), we can use the trigonometric identity:
sin(A) - sin(B) = 2 * cos((A + B)/2) * sin((A - B)/2)
Applying this identity to the given expression, we have:
sin(v+x) - sin(v-x) = 2 * cos((v+x + v-x)/2) * sin((v+x - v+x)/2)
= 2 * cos(v) * sin(x)
= 2cos(v)sin(x)
So, the simplified expression is 2cos(v)sin(x).
Now, let's rewrite the expression sin(2cos^(-1)x) as an algebraic expression in x.
To simplify this expression, we need to use the inverse cosine function, or arccosine, which is denoted as cos^(-1)x or acos(x). The range of the inverse cosine function is [0, π].
Using the double angle formula for sine, sin(2θ) = 2sin(θ)cos(θ), we can rewrite the given expression as:
sin(2cos^(-1)x) = 2sin(cos^(-1)x)cos(cos^(-1)x)
Now, we use the relationship between sine and cosine in a right triangle:
sin(θ) = opposite/hypotenuse
cos(θ) = adjacent/hypotenuse
Let's assume that θ = cos^(-1)x.
For a right triangle with an angle θ = cos^(-1)x, the adjacent side is 1, and the hypotenuse is x. Therefore, the opposite side is √(x^2 - 1).
Using these values, we have:
sin(2cos^(-1)x) = 2sin(cos^(-1)x)cos(cos^(-1)x)
= 2(sin(θ))(cos(θ))
= 2(√(x^2 - 1))/x * 1/x
= 2(√(x^2 - 1))/x^2
Thus, the algebraic expression for sin(2cos^(-1)x) is 2(√(x^2 - 1))/x^2.