sinx = 4/5 and x terminates in Quadrant II
Find sin2x and cos2x
How to get the answers, which are sin2x = -24/25, cos2x = -7/25?
Using the double angle formula, we can find sin2x and cos2x:
sin2x = 2sinxcosx
cos2x = cos2x - sin2x
Therefore,
sin2x = 2(4/5)(-3/5) = -24/25
cos2x = (4/5)^2 - (-24/25)^2 = -7/25
To find sin2x and cos2x, we'll need to use some trigonometric identities and properties. Here's how you can get the answers:
1. Start with the given information: sin(x) = 4/5 and x terminates in Quadrant II.
2. Since sin(x) = 4/5 and x is in Quadrant II, we know that the cosine of x will be negative. We can find this using the Pythagorean Identity: cos(x) = √(1 - sin^2(x)), which gives: cos(x) = √(1 - (4/5)^2) = √(1 - 16/25) = √(9/25) = 3/5.
3. Now, we can use the double-angle formulas to find sin2x and cos2x:
a) sin2x = 2sin(x)cos(x)
= 2 * (4/5) * (3/5)
= 24/25
b) cos2x = cos^2(x) - sin^2(x)
= (3/5)^2 - (4/5)^2
= 9/25 - 16/25
= -7/25
Therefore, the answers are sin2x = -24/25 and cos2x = -7/25.
To find the values of sin(2x) and cos(2x) given that sin(x) = 4/5 and x terminates in Quadrant II, follow these steps:
Step 1: Determine the value of cos(x).
Since sin(x) = 4/5, we can use the Pythagorean identity sin^2(x) + cos^2(x) = 1 to find the value of cos(x).
Start by substituting sin(x) with its value: (4/5)^2 + cos^2(x) = 1.
Simplifying the equation gives 16/25 + cos^2(x) = 1.
To isolate cos^2(x), subtract 16/25 from both sides: cos^2(x) = 9/25.
Finally, take the square root of both sides to find the value of cos(x): cos(x) = ±3/5.
Since x terminates in Quadrant II, the value of cos(x) is negative. Therefore, cos(x) = -3/5.
Step 2: Use the double-angle formulas.
sin(2x) = 2sin(x)cos(x) and cos(2x) = cos^2(x) - sin^2(x).
Substitute the known values for sin(x) and cos(x):
sin(2x) = 2(4/5)(-3/5) = -24/25.
cos(2x) = (-3/5)^2 - (4/5)^2 = 9/25 - 16/25 = -7/25.
Therefore, sin(2x) = -24/25 and cos(2x) = -7/25.