Help with Trig equation and all solutions

[futuremathwiz]

Hello, can someone please explain why were adding just pi and not two pi to this trig equation. I kept reading my textbook and everywhere online and they all say cos and sin have a periodicity of two pi while tan has a periodicity of just pi. Im sorta just lost now on Question 33.

 

[MarkFL]

Hello, and welcome to FMH!

Equation 33 implies:

[MATH]\cos(\theta)=\pm\frac{1}{\sqrt{2}}[/MATH]
I would write the solution as:

[MATH]\theta=\frac{\pi}{4}+\frac{\pi}{2}k=\frac{\pi}{4}(2k+1)[/MATH]
This is equivalent to the solution given in your image, but more succinctly written. There is a solution in all 4 quadrants because of both signs being present when we take the square root of the cosine function.

 

[Dr.Peterson]

Hello, can someone please explain why were adding just pi and not two pi to this trig equation. I kept reading my textbook and everywhere online and they all say cos and sin have a periodicity of two pi while tan has a periodicity of just pi. Im sorta just lost now on Question 33.

It sounds like you are trying to work backward from the solution, and you think the [MATH]+k\pi[/MATH] would be due to the period being [MATH]\pi[/MATH]. That's a dangerous thing to do; there is no such period involved here.

I hope you have read MarkFL's answer, and worked along with him to see where the answer comes from:

There are two values for the cosine: (a) [MATH]\cos(\theta) = \frac{1}{\sqrt{2}}[/MATH] and (b) [MATH]\cos(\theta) = -\frac{1}{\sqrt{2}}[/MATH].

Each of these equations has two solutions within one cycle: (a) [MATH]\theta = \frac{\pi}{4}[/MATH] or [MATH]\theta = \frac{7\pi}{4}[/MATH]; (b) [MATH]\theta = \frac{3\pi}{4}[/MATH] or [MATH]\theta = \frac{5\pi}{4}[/MATH]. Both MarkFL's answer and the book's answer are attempts to combine these into one or two expressions (adding [MATH]+2k\pi[/MATH] to cover coterminal angles).