if cos 2 theta = -(1/3) and theta is in Quadrant 2, find sin theta, cos theta, tan theta, and draw triangle theta
-1/3 = cos 2theta = 1 - 2 sin^2 theta
2 sin^2 theta = 4/3
sin theta = sqrt(2/3)
cos theta = sqrt(1 - 2/3) = -sqrt(1/3)
(minus since theta must be in quadrant 2)
tan theta = -sqrt2
theta = 125.26 degrees
You draw the figure
To solve this question step-by-step, let's start with the given information:
1. cos(2θ) = -1/3
2. Theta (θ) is in Quadrant 2
Step 1: Use the double-angle formula for cosine to find cos(θ):
The double-angle formula for cosine is:
cos(2θ) = 2cos²(θ) - 1
Substituting the given value, we have:
-1/3 = 2cos²(θ) - 1
Rearranging the equation:
2cos²(θ) = -1/3 + 1
2cos²(θ) = 2/3
Divide both sides by 2:
cos²(θ) = 1/3
Taking the square root of both sides:
cos(θ) = ±√(1/3)
Since we know that θ is in Quadrant 2, the value of cos(θ) should be negative:
cos(θ) = -√(1/3)
Step 2: Find sin(θ) using the Pythagorean identity:
The Pythagorean identity for sine and cosine states that:
sin²(θ) + cos²(θ) = 1
Substituting the value of cos(θ) we found earlier:
sin²(θ) + (-√(1/3))² = 1
sin²(θ) + 1/3 = 1
Rearranging the equation:
sin²(θ) = 1 - 1/3
sin²(θ) = 2/3
Taking the square root of both sides:
sin(θ) = ±√(2/3)
Since we know that θ is in Quadrant 2, the value of sin(θ) should be positive:
sin(θ) = √(2/3)
Step 3: Find tan(θ) using the definition of tangent:
tan(θ) = sin(θ) / cos(θ)
Substituting the values we found earlier:
tan(θ) = √(2/3) / -√(1/3)
Simplifying the expression by multiplying the numerator and denominator by √3:
tan(θ) = (√(2/3) * √3) / (-√(1/3) * √3)
tan(θ) = √(6/9) / -√(3/3)
tan(θ) = √(6/9) / -1
tan(θ) = -√(6/9)
Step 4: Draw the triangle θ in Quadrant 2:
In Quadrant 2, the x-coordinate (cosine) is negative, and the y-coordinate (sine) is positive. Since we found that cos(θ) = -√(1/3) and sin(θ) = √(2/3), we can draw a triangle with a hypotenuse of length 1 (for simplicity), a base of -√(1/3), and a height of √(2/3).
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θ -√(1/3)
This triangle represents θ in Quadrant 2 with a hypotenuse of 1, a base of -√(1/3), and a height of √(2/3).
To find the values of sin theta, cos theta, and tan theta, we can use the given value of cos 2 theta and apply the double-angle formula for cosine. Since theta is in Quadrant 2, we know that cosine is negative and sine is positive in this quadrant.
Step 1: Find cos theta
We know that cos 2 theta = -(1/3). To apply the double-angle formula, we can write it as:
cos 2 theta = 2 cos^2 theta - 1
Substituting the given value, we have:
-(1/3) = 2 cos^2 theta - 1
Simplifying the equation, we get:
2 cos^2 theta = 2/3
cos^2 theta = 1/3
cos theta = sqrt(1/3) or -sqrt(1/3)
Since theta is in Quadrant 2 (cos is negative), the value of cos theta is -sqrt(1/3).
Step 2: Find sin theta
Since cos theta = -sqrt(1/3), we can use the Pythagorean identity: sin^2 theta + cos^2 theta = 1.
substituting the value of cos theta, we have:
sin^2 theta + (-sqrt(1/3))^2 = 1
sin^2 theta + 1/3 = 1
sin^2 theta = 2/3
sin theta = sqrt(2/3)
Since theta is in Quadrant 2 (sin is positive), the value of sin theta is sqrt(2/3).
Step 3: Find tan theta
To find tan theta, we can use the relationship: tan theta = sin theta / cos theta.
Substituting the values of sin theta and cos theta, we have:
tan theta = (sqrt(2/3)) / (-sqrt(1/3))
tan theta = -sqrt(2)
Therefore, the value of tan theta is -sqrt(2).
Step 4: Drawing the triangle theta
To draw the triangle theta, we can use the known values of sin theta and cos theta. Since cos theta is negative and sin theta is positive in Quadrant 2, the triangle will be drawn with the hypotenuse in the second quadrant and the opposite side above the x-axis.
Here's an approximate representation of the triangle theta:
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In this triangle, the side adjacent to theta is -sqrt(1/3), the side opposite to theta is sqrt(2/3), and the hypotenuse is 1.