Find all the angles θ in the interval [-2pi, 2pi] for which cos(3θ) = 1/sqrt(2)
I'm not sure why I would have to look for angles in 3 revolutions of the circle (3 positive, 3 negative).
Also how would I convert cos(3θ) = 1/sqrt(2) to a radian value? Do I just need to know the unit circle by heart?
You should know, or by use of your calculator, that cos(±45°) = 1/√2
so 3θ = ±45° and θ = ±15°
now the period of cos(3θ) is 360°/3 or 120°
so by repeatedly adding and subtracting 120° to any answer will yield a new answer
Your domain is -360° to 360°, so answers in degrees are:
±15, ±135, ±255, ±105, ±225, and ±345
in radians, you know that π/6 = 30°
so 15° = π/12
to quickly convert the above degrees to radians I do the following steps
135/15 = 9
then 135° = 9(π/12) = 3π/4
continuing in this way, the radian answers are:
±π/12, ±3π/4, ±17π/12, ±7π/12, ±5π/4, and ±23π/12
Proof:
www.wolframalpha.com/input/?i=plot+y+%3D+cos(3x)+,+y+%3D+1%2F%E2%88%9A2+from+-2%CF%80+to+2%CF%80
The above post was my me, forgot to put in my name
Page 2 of the following has a nice unit circle,
have it handy
tutorial.math.lamar.edu/pdf/trig_cheat_sheet_reduced.pdf
Thanks. I'm not permitted to use a calculator, which makes this a lot harder.
To solve the equation cos(3θ) = 1/√2, you're correct that it's not necessary to look for angles in 3 revolutions of the circle initially. However, if you're searching for all possible solutions in the given interval [-2π, 2π], it's important to consider multiple revolutions of the unit circle.
Now, let's break down the process of finding the angles θ step by step:
1. Start by finding the principal angle:
To solve for θ, we need to solve cos(3θ) = 1/√2.
Using the inverse cosine function, we obtain:
3θ = arccos(1/√2)
2. Find the principal value of θ:
Divide both sides of the equation by 3:
θ = (1/3) * arccos(1/√2)
3. Convert the principal value to radian form:
To get the angle in radians, it's not necessary to memorize the entire unit circle. Instead, we can use the basic or special angles along with trigonometric identities.
In this case, cos(π/4) = 1/√2. So, we can express the principal value in terms of the special angle as:
θ = (1/3) * (π/4) [Since arccos(1/√2) = π/4]
4. Evaluate the principal solution:
Simplify the expression:
θ = π/12
5. Find additional solutions:
To find other solutions in the given interval [-2π, 2π], we need to add or subtract multiples of the period.
Since cos(θ) has a period of 2π, the solutions are obtained by adding or subtracting multiples of 2π/3 to the principal solution.
Therefore, the additional solutions are:
θ = π/12 + 2πk/3 or θ = π/12 - 2πk/3 where k is an integer.
By following these steps, you can find all the angles θ in the interval [-2π, 2π] for which cos(3θ) = 1/√2.