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From what you posted, it appears that you are to simplify:
\(\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)\)
However, I suspect that you are instead being asked to simplify:
\(\displaystyle y=\sin\left(\cos^{-1}(x) \right)\)
Which is the case?
Okay, I just wanted to be sure.The question just states:
Simplify the following expression, draw a right triangle to assist you.
\(\displaystyle sin^{-1}\left(\cos(x) \right)\)
Okay, I just wanted to be sure.
What do we get if we take the sine of both sides of the equation:
\(\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)\)
and then what conclusion can we reach regarding the relationship between $\theta$ and $x$?
Simplify a trig expression using a right triangle
I have the expression \(\displaystyle sin^{-1}(cosx)\)
Mark is right: the identity $\sin^2\theta+\cos^2\theta=1$ (which comes from the Pythagorean theorem) is used to simplify \(\displaystyle \sin(\cos^{-1}(x))\), not \(\displaystyle \sin^{-1}\left(\cos(x) \right)\). If the question does mention the right triangle, it may very well have a typo. For the latter expression, it is enough to note that $\cos(x)=\sin(\pi/2-x)$ for all $x$. However, what follows requires some tinkering to get an expression for all $x$ because $\sin^{-1}(\sin(x))=x$ only for $-\pi/2\le x\le\pi/2$. I can show the function that works for all $x$ if needed.From what you posted, it appears that you are to simplify:
\(\displaystyle \theta=\sin^{-1}\left(\cos(x) \right)\)
However, I suspect that you are instead being asked to simplify:
\(\displaystyle y=\sin\left(\cos^{-1}(x) \right)\)
I took a picture of it out of my textbook to show you guys that is how it was presented.Mark is right: the identity $\sin^2\theta+\cos^2\theta=1$ (which comes from the Pythagorean theorem) is used to simplify \(\displaystyle \sin(\cos^{-1}(x))\), not \(\displaystyle \sin^{-1}\left(\cos(x) \right)\). If the question does mention the right triangle, it may very well have a typo. For the latter expression, it is enough to note that $\cos(x)=\sin(\pi/2-x)$ for all $x$. However, what follows requires some tinkering to get an expression for all $x$ because $\sin^{-1}(\sin(x))=x$ only for $-\pi/2\le x\le\pi/2$. I can show the function that works for all $x$ if needed.
Just for information, to simplify $\sin(\cos^{-1}(x))$ where $-1\le x\le 1$, assume, without loss of generality, that $\cos(\theta)=x$ for some $0\le\theta\le\pi$. Then $\theta=\cos^{-1}(x)$. What, then, is $\sin(\theta)$ (note that $\sin(\theta)\ge0$)?
That the angles should be equal?Because of the instruction to draw a triangle, this is why I thought my second interpretation in my first post was more likely the case. We really do not need to draw a triangle for this. But we could:
View attachment 1266
Do you see that:
\(\displaystyle \sin(\theta)=\cos(x)=\frac{a}{c}\) ?
If the sine of one angle is equal to the cosine of another angle, and given that the name cosine derives roughly from "COmpliment of SINE", what do we know about $\theta$ and $x$?
I appreciate the upload, like Evgeny.Makarov, I think the author of the problem has simply made a typo. However, the problem as given can be done, so perhaps this is actually what was intended.
Do you see what relationship must hold for $\theta$ and $x$, given that they are two angles in a right triangle?
edit: given that $\theta$ is used in the problem statement rather than $x$, simply reverse the two variables as I have used them above:
\(\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)\)
\(\displaystyle \sin(x)=\cos(\theta)\)
\(\displaystyle x+\theta=?\)
Yeah, I am in calculus right now. It's just my high school never did much with trigonometry unfortunately, so it's one of my weak areas I'm trying to catch up on.Yes, assuming you mean degrees :
\(\displaystyle x+\theta=90^{\circ}\)
or in radians (which I prefer, and if you plan on taking calculus you will want to get used to using radians):
\(\displaystyle x+\theta=\frac{\pi}{2}\)
So, we may solve for $x$ to get:
\(\displaystyle x=\frac{\pi}{2}-\theta\)
and since:
\(\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)\), we may then state:
\(\displaystyle \sin^{-1}\left(\cos(\theta) \right)=\frac{\pi}{2}-\theta\)