Now, I understand problems such as sin2theta=1 up until the point where I have to find all the solutions. For example, sin2theta=90, theta=45. I get that. But where do I go from there? If someone could explain in both degrees and radians (not using a calculator) that would be GREATLY appreciated. Thank you.
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Need help on a trig problem
- Thread starter Faith54
- Start date
Hello, Faith54!
For example: if \(\displaystyle \sin\theta\,=\,\frac{1}{2}\), we might conclude: \(\displaystyle \theta\,=\,30^o\) . . . and stop.
But there many other angles which satisfy the equation:
. . \(\displaystyle 150^o,\,390^o,\,510^o,\,\cdots\;\;\) and: -\(\displaystyle 210^o,\,\)-\(\displaystyle 330^o,\,\)-\(\displaystyle 570^o,\,\)-\(\displaystyle 690^o,\,\cdots\)
So how do we include all of them?
For this example, there are two basic answers:
. . \(\displaystyle \theta\,=\,30^o\) in quadrant 1, and \(\displaystyle \theta\,=\,150^o\) in quadrant 2.
All the others are simply: \(\displaystyle \;\text{(the basic angle)}\,\pm\,\text{(a multiple of 360}^o)\)
The solutions can be written: \(\displaystyle \;\theta\:=\:\begin{Bmatrix}30^o\,+\,n\cdot360^o \\ 150^o\,+\,n\cdot360^o\end{Bmatrix}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Back to your problem: \(\displaystyle \;\sin(2\theta)\:=\:1\)
The basic answer is: \(\displaystyle \:2\theta\:=\:90^o\)
The general answer is: \(\displaystyle \:2\theta\:=\:90^o\,+\,n\cdot360^o\;\) for any integer \(\displaystyle n\).
So we have: \(\displaystyle \;2\theta\;=\;\cdots\,-630^o,\.-270^o,\,90^o,\,450^o, \,810^o,\,\cdots\)
Therefore: \(\displaystyle \;\theta\;=\;\cdots\,-315^o,\,-135^o,\,45^o,\,225^o,
\,810^o,\,\cdots\)
If the problem restricts \(\displaystyle \theta\) to, say, \(\displaystyle 0^o\,\leq\,\theta\,<\,360^o\)
. . we just pick the ones in that interval: \(\displaystyle \;\theta\,=\,45^o,\,225^o\)
The problem is that inverse trig functions have multiple answers.Now, I understand problems such as: \(\displaystyle \sin(2\theta)\,=\,1\)
up until the point where I have to find all the solutions.
For example: \(\displaystyle 2\theta\,=\,90^o\;\;\Rightarrow\;\;\theta\,=\,45^o\)
I get that.. But where do I go from there?
For example: if \(\displaystyle \sin\theta\,=\,\frac{1}{2}\), we might conclude: \(\displaystyle \theta\,=\,30^o\) . . . and stop.
But there many other angles which satisfy the equation:
. . \(\displaystyle 150^o,\,390^o,\,510^o,\,\cdots\;\;\) and: -\(\displaystyle 210^o,\,\)-\(\displaystyle 330^o,\,\)-\(\displaystyle 570^o,\,\)-\(\displaystyle 690^o,\,\cdots\)
So how do we include all of them?
For this example, there are two basic answers:
. . \(\displaystyle \theta\,=\,30^o\) in quadrant 1, and \(\displaystyle \theta\,=\,150^o\) in quadrant 2.
All the others are simply: \(\displaystyle \;\text{(the basic angle)}\,\pm\,\text{(a multiple of 360}^o)\)
The solutions can be written: \(\displaystyle \;\theta\:=\:\begin{Bmatrix}30^o\,+\,n\cdot360^o \\ 150^o\,+\,n\cdot360^o\end{Bmatrix}\)
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
Back to your problem: \(\displaystyle \;\sin(2\theta)\:=\:1\)
The basic answer is: \(\displaystyle \:2\theta\:=\:90^o\)
The general answer is: \(\displaystyle \:2\theta\:=\:90^o\,+\,n\cdot360^o\;\) for any integer \(\displaystyle n\).
So we have: \(\displaystyle \;2\theta\;=\;\cdots\,-630^o,\.-270^o,\,90^o,\,450^o, \,810^o,\,\cdots\)
Therefore: \(\displaystyle \;\theta\;=\;\cdots\,-315^o,\,-135^o,\,45^o,\,225^o,
\,810^o,\,\cdots\)
If the problem restricts \(\displaystyle \theta\) to, say, \(\displaystyle 0^o\,\leq\,\theta\,<\,360^o\)
. . we just pick the ones in that interval: \(\displaystyle \;\theta\,=\,45^o,\,225^o\)