Simplify a trig. expression using a right triangle

[bnosam]

I have the expression \(\displaystyle sin^{-1}(cosx)\)

I'm not sure how to simplify this at all. I've never done a problem like this and it's in my textbook as a review question.

A quick boot in the right direction would help

 

[MarkFL]

Re: Simplify a trig expression using a right triangle

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?

 

[bnosam]

Re: Simplify a trig expression using a right triangle

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?

The question just states:
Simplify the following expression, draw a right triangle to assist you.

\(\displaystyle sin^{-1}\left(\cos(x) \right)\)

 

[MarkFL]

Re: Simplify a trig expression using a right triangle

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. (Sun)

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 $x$?

 

[bnosam]

Re: Simplify a trig expression using a right triangle

Okay, I just wanted to be sure. (Sun)

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 $x$?

\(\displaystyle sin(\theta) = cos(x) \)

I'm not quite sure how drawing a triangle would fit into this so far.

 

[Evgeny.Makarov]

Re: Simplify a trig expression using a right triangle

Simplify a trig expression using a right triangle

I have the expression \(\displaystyle sin^{-1}(cosx)\)

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:

\(\displaystyle y=\sin\left(\cos^{-1}(x) \right)\)

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 $\sin(\theta)\ge0$)?

 

[MarkFL]

Re: Simplify a trig expression using a right triangle

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 $x$?

 

[bnosam]

Re: Simplify a trig expression using a right triangle

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 $\sin(\theta)\ge0$)?

I took a picture of it out of my textbook to show you guys that is how it was presented.

- - - Updated - - -

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:

https://www.physicsforums.com/attachments/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 $x$?

That the angles should be equal?

 

[MarkFL]

Re: Simplify a trig expression using a right triangle

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:

\(\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)\)

\(\displaystyle \sin(x)=\cos(\theta)\)

\(\displaystyle x+\theta=?\)

 

[bnosam]

Re: Simplify a trig expression using a right triangle

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:

\(\displaystyle x=\sin^{-1}\left(\cos(\theta) \right)\)

\(\displaystyle \sin(x)=\cos(\theta)\)

\(\displaystyle x+\theta=?\)

\(\displaystyle x+\theta= 90\)

 

[MarkFL]

Re: Simplify a trig expression using a right triangle

Yes, assuming you mean degrees (Tongueout):

\(\displaystyle x+\theta=90^{\circ}\)

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:

\(\displaystyle \sin^{-1}\left(\cos(\theta) \right)=\frac{\pi}{2}-\theta\)

 

[bnosam]

Re: Simplify a trig expression using a right triangle

Yes, assuming you mean degrees (Tongueout):

\(\displaystyle x+\theta=90^{\circ}\)

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:

\(\displaystyle \sin^{-1}\left(\cos(\theta) \right)=\frac{\pi}{2}-\theta\)

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.

Thanks for your help.

 

[MarkFL]

Re: Simplify a trig expression using a right triangle

Okay, good deal!

You are wise to work on strengthening your trig. skills, as it is use extensively in the study of calculus. I hope it goes well, and we are here to help you!(Sun)