# TrigonometrySimplify a trig. expression using a right triangle

#### bnosam

##### New member
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

Last edited:

#### MarkFL

Staff member
Re: Simplify a trig expression using a right triangle

From what you posted, it appears that you are to simplify:

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

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

Which is the case?

#### bnosam

##### New member
Re: Simplify a trig expression using a right triangle

From what you posted, it appears that you are to simplify:

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

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

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

Staff member
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.

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

##### New member
Re: Simplify a trig expression using a right triangle

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 $x$?

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

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

#### Evgeny.Makarov

##### Well-known member
MHB Math Scholar
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:

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

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

Staff member
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:

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

##### New member
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:

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 $x$?
That the angles should be equal?

#### Attachments

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#### MarkFL

Staff member
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

##### New member
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

Staff member
Re: Simplify a trig expression using a right triangle

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:

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

#### bnosam

##### New member
Re: Simplify a trig expression using a right triangle

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:

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