find sin2x, cos2x, and tan2x if sinx= -2/sqrt 5 and x terminates in quadrant III

Well, since sin(x) = -2/√5, we know that the sine of x in quadrant III is negative. This means that cos(x) in quadrant III will also be negative.

To find sin(2x), cos(2x), and tan(2x), we can use the double angle formulas. The double angle formulas are:

sin(2x) = 2sin(x)cos(x)
cos(2x) = cos^2(x) - sin^2(x)
tan(2x) = 2tan(x) / (1 - tan^2(x))

Let's start by finding cos(x). Since sin(x) = -2/√5, we can use the Pythagorean identity to find cos(x):

cos^2(x) + sin^2(x) = 1
cos^2(x) + (-2/√5)^2 = 1
cos^2(x) + 4/5 = 1
cos^2(x) = 1 - 4/5
cos^2(x) = 1/5

Since cos(x) is negative in quadrant III, we take the negative square root:

cos(x) = -√(1/5)

Now, let's find sin(2x):

sin(2x) = 2sin(x)cos(x)
sin(2x) = 2(-2/√5)(-√(1/5))
sin(2x) = 4/√5

Next, let's find cos(2x):

cos(2x) = cos^2(x) - sin^2(x)
cos(2x) = (-√(1/5))^2 - (-2/√5)^2
cos(2x) = 1/5 - 4/5
cos(2x) = -3/5

Finally, let's find tan(2x):

tan(2x) = 2tan(x) / (1 - tan^2(x))
tan(2x) = 2(-2/√5) / (1 - (-2/√5)^2)
tan(2x) = -4√5 / (1 - 4/5)
tan(2x) = -4√5 / (1/5)
tan(2x) = -20√5

So, sin2x = 4/√5, cos2x = -3/5, and tan2x = -20√5.

To find sin(2x), cos(2x), and tan(2x), we can use the double angle identities.

Let's start by finding sin(2x). The double angle identity for sine is:

sin(2x) = 2 * sin(x) * cos(x)

We already know sin(x) = -2/√5. To find cos(x), we can use the Pythagorean identity:

sin^2(x) + cos^2(x) = 1

((-2/√5)^2) + cos^2(x) = 1
4/5 + cos^2(x) = 1
cos^2(x) = 1 - 4/5
cos^2(x) = 1/5

Taking the square root of both sides, we get:

cos(x) = ±1/√5

Since x is in quadrant III, where cosine is negative, we take the negative root:

cos(x) = -1/√5

Now we can substitute these values into the double angle formula for sine:

sin(2x) = 2 * (-2/√5) * (-1/√5)
sin(2x) = 4/5

So, sin(2x) = 4/5.

Next, let's find cos(2x). The double angle identity for cosine is:

cos(2x) = cos^2(x) - sin^2(x)

We already know cos(x) = -1/√5 and sin(x) = -2/√5. Substituting these values into the formula, we have:

cos(2x) = (-1/√5)^2 - (-2/√5)^2
cos(2x) = 1/5 - 4/5
cos(2x) = -3/5

So, cos(2x) = -3/5.

Finally, let's find tan(2x). The double angle identity for tangent is:

tan(2x) = (2 * tan(x))/(1 - tan^2(x))

We know tan(x) = sin(x)/cos(x), so:

tan(x) = (-2/√5)/(-1/√5)
tan(x) = 2

Substituting this into the formula for tan(2x), we have:

tan(2x) = (2 * 2)/(1 - 2^2)
tan(2x) = 4/(-3)
tan(2x) = -4/3

Therefore, tan(2x) = -4/3.

To find the values of sin2x, cos2x, and tan2x, we first need to find the values of sinx, cosx, and tanx.

Given that sinx = -2/√5, and x terminates in Quadrant III, we can determine the values of sinx, cosx, and tanx as follows:

Since sinx is negative in Quadrant III, we know that cosx must be positive. Using the Pythagorean identity sin^2x + cos^2x = 1, we can solve for cosx.

sin^2x + cos^2x = 1
(-2/√5)^2 + cos^2x = 1
4/5 + cos^2x = 1
cos^2x = 1 - 4/5
cos^2x = 1/5

Taking the square root of both sides, we get:

cosx = ±√(1/5)

Since cosx is positive in Quadrant III, we take the positive square root:

cosx = √(1/5)

Now that we have the values of sinx and cosx, we can find the value of tanx using the equation:

tanx = sinx / cosx

tanx = (-2/√5) / √(1/5)
tanx = -2

Now let's find sin2x:

sin2x = 2 * sinx * cosx

sin2x = 2 * (-2/√5) * √(1/5)
sin2x = -4/√5

For cos2x, we can use the double-angle formula:

cos2x = cos^2x - sin^2x

cos2x = (1/5) - [(-2/√5)^2]
cos2x = 1/5 - 4/5
cos2x = -3/5

Finally, let's calculate tan2x using the double-angle formula:

tan2x = (2 * tanx) / (1 - tan^2x)

tan2x = (2 * -2) / (1 - (-2)^2)
tan2x = -4 / (1 - 4)
tan2x = 4/3

In summary, the values of sin2x, cos2x, and tan2x, given sinx = -2/√5 and x terminates in Quadrant III, are:

sin2x = -4/√5
cos2x = -3/5
tan2x = 4/3

if sinx = -2/√5 in III

y = -2, r = √5 , by Pythagoras,
x^2 + 4 = 5
x = -1 in III

cosx = -1/√5
tanx = sinx/cosx = (-2/√)/(-1/√5) = 2

sin 2x = 2sinxcosx = 2(-2/√5)(-1/√5) = 4/5
cos 2x = cos^2 x - sin^2 x = 1/5 - 4/5 = -3/5
tan 2x
= 2tanx/(1 - tan^2 x)
=2(2)/(1 - 4) = -4/3