If sin theta = x and -pi/2 is less than or equal to theta less than or equal to pi/2, find sin 2theta and cos 2theta in terms of x.
Is it alright if I recommend you to my classmates tommorow? YOu seem to know more than my professor!
follow the two formulas I pointed out to you in
http://www.jiskha.com/display.cgi?id=1209674791
Many of us here are retired teachers or profs.
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Thank you for the compliment! I'm glad to be of help. Yes, it is absolutely alright to recommend me to your classmates. I'm here to assist with any questions or problems they might have.
Now, let's move on to your question. We are given that sin(theta) = x, where -pi/2 ≤ theta ≤ pi/2. We need to find sin(2theta) and cos(2theta) in terms of x.
To solve this, we'll use the following trigonometric identities:
1) sin(2theta) = 2sin(theta)cos(theta)
2) cos(2theta) = cos^2(theta) - sin^2(theta)
First, let's find cos(theta). We know that sin^2(theta) + cos^2(theta) = 1 (from the Pythagorean identity). Since we are given sin(theta) = x, we can substitute it into the equation:
x^2 + cos^2(theta) = 1
Rearranging the equation, we have:
cos^2(theta) = 1 - x^2
Taking the square root of both sides, we get:
cos(theta) = ±√(1 - x^2)
But since -π/2 ≤ theta ≤ π/2, we know that cos(theta) is positive in this range. Therefore:
cos(theta) = √(1 - x^2)
Now we can substitute sin(theta) and cos(theta) into the trigonometric identities to find sin(2theta) and cos(2theta):
1) sin(2theta) = 2sin(theta)cos(theta)
Substituting sin(theta) = x and cos(theta) = √(1 - x^2), we get:
sin(2theta) = 2x√(1 - x^2)
2) cos(2theta) = cos^2(theta) - sin^2(theta)
Substituting sin(theta) = x and cos(theta) = √(1 - x^2), we get:
cos(2theta) = (1 - x^2) - x^2
Simplifying:
cos(2theta) = 1 - 2x^2
So, sin(2theta) = 2x√(1 - x^2) and cos(2theta) = 1 - 2x^2.
I hope that helps! Let me know if you have any more questions.