Find sin 2x if sinx=3/5 and x is in Quadrant II.
sinx = 3/5 = opposite/hypotenuse = y/r
and you are in quad II
make a sketch, you should recognize the 3-4-5 Pythagorean triangle, if not ...
x^2 + y^2 = r^2
x^2 + 9 = 25
x = ± 4, but we are in II , so x = -4
cosx = -4/5
sin (2x) = 2sinxcosx = 2(3/5)(-4/5) = -24/25
check it with a calculator to satisfy your curiosity.
Well, well, well, we have a Quadrant II situation here! So, we know that sin x is positive in Quadrant II. We also have the information that sin x equals 3/5. Now, let's find sin 2x!
But before that, let me remind you of a little trigonometric identity called the double-angle identity for sin:
sin 2x = 2sin x cos x
Okay, now let's get down to business. We need to find cos x because we have sin x. Since sin x is positive in Quadrant II, this means that cos x is negative. To find cos x, we can use the Pythagorean identity:
sin² x + cos² x = 1
Now, substitute the value of sin x (which is 3/5):
(3/5)² + cos² x = 1
Simplifying that equation, we get:
9/25 + cos² x = 1
Subtracting 9/25 from both sides, we have:
cos² x = 16/25
Now, taking the square root of both sides (and remember, we want the negative square root since cos x is negative in Quadrant II), we get:
cos x = -4/5
Alrighty then! We have sin x = 3/5 and cos x = -4/5. To find sin 2x, we can apply the double-angle identity for sin:
sin 2x = 2sin x cos x
Plugging in the values, we get:
sin 2x = 2(3/5)(-4/5)
Simplifying that, we get:
sin 2x = -24/25
So, sin 2x is equal to -24/25. Voila! Keep rocking that Quadrant II knowledge!
To find sin 2x, we can use the double angle formula for sine:
sin 2x = 2sin x cos x
We are given that sin x = 3/5 and x is in Quadrant II.
In Quadrant II, sine is positive, while cosine is negative.
Since sin x = 3/5, we can determine the value of cos x using the Pythagorean identity:
cos^2 x = 1 - sin^2 x
cos^2 x = 1 - (3/5)^2
cos^2 x = 1 - 9/25
cos^2 x = 16/25
cos x = √(16/25)
cos x = 4/5
Now we can substitute the values of sin x and cos x into the double angle formula:
sin 2x = 2(sin x)(cos x)
sin 2x = 2(3/5)(4/5)
sin 2x = (2)(3)(4)/(5)(5)
sin 2x = 24/25
Therefore, sin 2x = 24/25 when sin x = 3/5 and x is in Quadrant II.
To find sin 2x, we can use the double angle identities for sine: sin 2x = 2sinxcosx.
Since we are given that sinx = 3/5 and x is in Quadrant II, we know that sinx is positive and cosx is negative in this quadrant.
First, we need to find cosx. To do that, we can use the Pythagorean identity: sin^2x + cos^2x = 1.
Given sinx = 3/5, we can square it: sin^2x = (3/5)^2 = 9/25.
Using the Pythagorean identity, we can solve for cos^2x: cos^2x = 1 - sin^2x = 1 - 9/25 = 16/25.
Since x is in Quadrant II, cosx is negative, so cosx = -sqrt(16/25) = -4/5.
Now, we can use the double angle identity to find sin 2x:
sin 2x = 2sinxcosx = 2 * (3/5) * (-4/5) = -24/25.
Therefore, sin 2x equals -24/25.