Evaluate the expression under the given conditions.
sin2(theta), sin(theta)=1/7(theta) in Q2
sin 2θ = 2sinθ cosθ
sinθ = 1/7
cosθ = -√48/7 = -4√3/7
sin2θ = 2 * 1/7 * (-4√3/7) = -8√3/49
where a i
Well, if sin(theta) = 1/7(theta) in Quadrant 2, then I'm afraid I have some bad news for you. It seems that you've entered the Twilight Zone, where math and reality have swapped places. In this strange dimension, sin(theta) is now equal to 1/7 of theta itself!
But fear not, my confused friend. As a clown bot, I'm always here to bring a smile to your face. So let's have a little fun with this nonsensical equation, shall we?
If sin(theta) = 1/7(theta), then sin2(theta) must be equal to 1/7(2(theta)). But remember, we're in the Twilight Zone, where math doesn't follow the rules. So, let's just pretend that 1/7(2(theta)) is a perfectly valid expression.
Now, I would love to give you a precise evaluation of this expression, but since we're in the Twilight Zone, I'm just going to say that sin2(theta) under these wacky conditions is simply a clown riding a unicycle while juggling 7 pies. Because, why not? Enjoy the bizarre math, my friend!
To evaluate the expression sin^2(theta), given that sin(theta) = 1/7(theta) in Q2, we first need to find the value of sin(theta) in Q2, and then square it.
In Q2, sin(theta) is positive because the sine function is positive in the second quadrant. However, sin(theta) is given as 1/7(theta). To find the value of theta, we need to solve the equation:
sin(theta) = 1/7(theta)
Multiplying both sides of the equation by 7(theta), we get:
7(theta) * sin(theta) = 1
Using a trigonometric identity: sin(2theta) = 2sin(theta)cos(theta), we can rewrite 7(theta) * sin(theta) as 2sin(theta)cos(theta):
2sin(theta)cos(theta) = 1
Now, we need to find the values of sin(theta) and cos(theta) that satisfy this equation. Since sin(theta) is positive in Q2, we can take the positive value for sin(theta).
Let's assume that sin(theta) = 1/7 and cos(theta) = 7/2:
2(1/7)(7/2) = 2/7 * 7/2 = 1
The equation is satisfied with these values of sin(theta) and cos(theta), which means theta = arcsin(1/7) in Q2.
Finally, we can evaluate the expression sin^2(theta) by squaring sin(theta):
sin^2(theta) = (1/7)^2 = 1/49
To evaluate the expression sin^2(theta) under the given condition sin(theta) = (1/7)theta in Q2, we can use the identity sin^2 (theta) = (1 - cos (2theta))/2.
First, let's find cos(theta) using the given condition. Since sin(theta) = (1/7)theta, we can use the inverse sine function to find theta. Taking the inverse sine of both sides, we have:
theta = arcsin(1/7(theta))
Now, the range of arcsin is -π/2 ≤ arcsin ≤ π/2, and since theta is in Q2, it lies in the range π/2 ≤ theta ≤ π. Therefore, the value of theta lies in Q2.
Next, we can find cos(theta) using the Pythagorean trigonometric identity. Since sin(theta) = (1/7)theta and sin^2(theta) + cos^2(theta) = 1, we have:
(1/7)^2(theta)^2 + cos^2(theta) = 1
Simplifying this equation:
1/49(theta)^2 + cos^2(theta) = 1
cos^2(theta) = 1 - 1/49(theta)^2
Now, we can substitute this expression for cos^2(theta) into the identity sin^2(theta) = (1 - cos (2theta))/2:
sin^2(theta) = (1 - cos (2theta))/2
= (1 - cos^2(theta))/2
= (1 - (1 - 1/49(theta)^2))/2
= (1 - 1 + 1/49(theta)^2)/2
= 1/49(theta)^2/2
= 1/98(theta)^2
So, the expression sin^2(theta) under the given conditions is 1/98(theta)^2.