e.g. tan 315 = -1, does that work in our equation ??
Yes, thank you again Reiny!
My next question is to find sin theta and cos theta.
cos 2theta = 1/3 and pi less than or equal to 2theta less than or equal to 2pi
I hope Im not overburdening you with question:/
<<I hope Im not overburdening you with question:/>>
lol, I love trig and do these kind of questions for recreation.
for this one you will need to know
1. sin2A = 2sinAcosA, and
2. cos2A = cos^2 A - sin^2 A
or 2cos^2 A -1
or 1 - 2sin^2 A
I will do the first, you do the second
cos2A = 1-2sin^2A
1/3 = 1 - 2sin^2 A
2 sin^2 A = 1 - 1/3
sin^2 A = 1/3
sinA = ±1/√3 (I used A instead of theta)
Can we use both ± ?
2theta is in either I or IV
then theta is in either I or II, but the sine in both of these is positive, so
sinA = +1/√3 only.
you try the second one.
Not at all! I'm here to help. Let's solve your next question step by step.
Given: cos(2θ) = 1/3 and π ≤ 2θ ≤ 2π.
To find sinθ and cosθ, we can utilize the double-angle identities for cosine and sine.
Let's start with the double-angle identity for cosine:
cos(2θ) = 1/3
Using the double-angle identity: cos(2θ) = 2cos^2(θ) - 1
Substituting the given value: 2cos^2(θ) - 1 = 1/3
Rearranging the equation: 2cos^2(θ) = 4/3
Dividing both sides by 2: cos^2(θ) = 2/3
Now, let's use the Pythagorean identity to find sinθ:
sin^2(θ) = 1 - cos^2(θ)
Substituting the value of cos^2(θ) we found earlier: sin^2(θ) = 1 - 2/3
Simplifying the equation: sin^2(θ) = 1/3
To find sinθ and cosθ, we will take the square root of both sides:
sinθ = ± √(1/3)
cosθ = ± √(2/3)
Since the given restriction is π ≤ 2θ ≤ 2π, we need to determine the suitable sign (+/-) for sinθ and cosθ based on the quadrants in which 2θ lies.
For π ≤ 2θ ≤ 2π, we know that 2θ falls in the second quadrant (π < 2θ < (3/2)π), where sinθ is positive and cosθ is negative. Therefore:
sinθ = √(1/3) [positive]
cosθ = -√(2/3) [negative]
So, the values of sinθ and cosθ are √(1/3) and -√(2/3) respectively, for the given condition.
I hope that helps! Let me know if you have any further questions.