Maximal value of domain for a function by looking at inverse function.

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$$\sin(2(x-\pi/6))=\sin(2x - \pi/3)$$

$\sin$ has an inverse in the interval $-\pi/2$ to $\pi/2$. So $$-\pi/2 \le 2x-\pi/3 \le \pi/2 \\ -\pi/2+\pi/3\le 2x \le \pi/2 + \pi/3 \\ -\pi/12 \le x \le 5\pi/12$$ So $[-a, a]$ should fit into $[-\pi/12, 5\pi/12]$ So $$a = \pi/12$$

The following was my original answer and is wrong! Please disregard. I was not thinking right!

Note that $\sin x$ has an inverse in an interval of length $\pi$. So $\sin 2x$ must have an inverse over an interval of length $\pi/2$. So $a=\pi/4$.

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confused

Updated on August 01, 2022

• confused over 1 year

The function g:[–a,a]→ R, g(x)=sin(2(x-π/6))has an inverse function.The maximum possible value of a is:

From what I understand the domain of g(x) is the range of g'(x). So I would try to find the inverse equation, work out the range, and then use this information to find the value of a, in the domain.

but i'm stuck at finding the range.

note: if i enter π/2 the calc says undefined so i'm assuming the value of a is π/2.

the answer is out of: π/12, 1, π/6, π/4, π/2

• confused almost 10 years
I'm not really understanding what an interval length is.
• user44197 almost 10 years
• confused almost 10 years
Uh... where? is it?
• user44197 almost 10 years
sorry, I am a slow typist :)
• confused almost 10 years
i dont understand how you get pi/6 on both sides? Can you explain please? haha, it's okay, I'm slower :(
• user44197 almost 10 years
I got pi/6 because I wanted to write pi/3 but wrote pi/6. It is fixed.
• confused almost 10 years
oh great! so with these inverse intervals? Is there a set one for cos/sin and tan? Do you know a link that talks more about inverse for trig? Cause I had no idea sin was inverse in-between those two values. I'd like to see it as a drawing, so I know what's going on, but I dont know where to even begin looking.
• confused almost 10 years
You know how you would normally go about finding inverse domains and range? Like I had done in my working out? Well this clearly doesn't apply to this question. So I'm assuming for all similar trig/inverse questions i would go about it like you did. Is this correct to believe?
• user44197 almost 10 years
Yeah for trig functions, you have to look at the graph and decide. $\sin$ is 1-1 between $-pi/2$ and $\pi/2$. (also between $\pi/2$ and $\3 pi/2$). Cosine is 1-1 from zero to $\pi$ (and also $\pi$ to $2 \pi$). You just have to pick the right interval depending on the problem.
• confused almost 10 years
so instead of writing -pi/2 can i write 3pi/2? For cosine- does it matter if you write zero instead of 2pi when working it out... etc?
• user44197 almost 10 years
yes, you just have to pick the interval so it contains the value you want. In this case you want $2x-pi/3$ should contain zero since you are looking for $[-a,a]$
• confused almost 10 years
okay yes, i got it - so look at the domain, and make a judgement?