# Angle equations

I have no idea what to do at all for this question, we just learned trig values and I don't see what to do for this question. I think it has something to do with co functions (as in cosine, cotan, csc).

Find a solution to the equation $$\tan \alpha = \cot (\alpha + 10^{\circ}).$$

Assume that all angles in which an unknown appears are acute angles.

[Added later] I also need to solve: $\sec(3 \beta + 10) = \csc( \beta + 8)$ (angles are in degrees).

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Please don't post questions in the imperative ("Find", "Solve", "Prove", etc). It reads is if you were issuing orders or assigning problems to the readers in the site, and many find it annoying and impolite. Having been on the site for almost two months, you may have noticed that one gets much better feedback by asking focused questions rather than simply by quoting a homework assignment. What are you having trouble with, and why? –  Arturo Magidin Jun 5 '11 at 19:31
I have no idea what to do at all for this question, we just learned trig values and I don't see what to do for this question. I think it has something to do with co functions. –  Adam Jun 5 '11 at 19:39
@Adam: Then please write that in the post, instead of just posting as if you were issuing orders. –  Arturo Magidin Jun 5 '11 at 19:43
@Adam: Please refrain from using [homework] as a sole tag. Surely you know the topic the question is in, please add that to the tags list. –  Asaf Karagila Jun 5 '11 at 19:46
It won't let me add new tags. –  Adam Jun 5 '11 at 19:48

Update: as for the added equation

$$\sec (3\beta +10{{}^\circ})=\csc (\beta +8{{}^\circ})$$

you may use the identity (see below)

$$\csc (a)=\sec (90{{}^\circ}-a),$$

where in this case $a=\beta +8{{}^\circ}$. Then you have

$$\csc (\beta +8{{}^\circ})=\sec \left( 90-\beta -8{{}^\circ}\right)$$

and

$$\sec (3\beta +10{{}^\circ})=\sec \left( 90-\beta -8{{}^\circ}\right) .$$

A solution $\beta$ can be found by equating $3\beta +10{{}^\circ}$ and $90-\beta -8{{}^\circ}$, and solve for $\beta$.

Hint: use the following identity for the complementary angles $a$ and $90{{}^\circ}-a$. [Edit: I note that this idea is the same as Isaac's. Edit end] $$\begin{equation*} \cot a=\tan \left( 90{{}^\circ}-a\right)\qquad (\ast) \end{equation*}$$

to get $\alpha =40{{}^\circ}$.

Added 4: Here is how you can get $\alpha =40{{}^\circ}$.

$$\begin{eqnarray*}\tan \alpha &=&\cot \left( \alpha +10{{}^\circ}\right) =\tan \left( 90{{}^\circ}-\alpha -10{{}^\circ}\right) \\&& \\&\Rightarrow &\alpha =90{{}^\circ}-\alpha -10{{}^\circ}\Leftrightarrow \alpha =40{{}^\circ} \end{eqnarray*}$$

$$\begin{eqnarray*} \sin \left( \frac{\pi }{2}-a\right) &=&\cos a \\ \cos \left( \frac{\pi }{2}-a\right) &=&\sin a \\ \tan \left( \frac{\pi }{2}-a\right) &=&\cot a \\ \csc \left( \frac{\pi }{2}-a\right) &=&\sec a \\ \sec \left( \frac{\pi }{2}-a\right) &=&\csc a. \end{eqnarray*}$$

Added 2: Derivation of the identity $(\ast)$. Let $P(x,y)$ and $I(1,0,)$ be two points on the unit circle centered at $O(0,0)$. Denote $\measuredangle POI=a$. Let $P^{\prime }(x^{\prime},y^{\prime })$ be the point on the unit circle such that $\measuredangle P^{\prime }OI=\frac{\pi }{2}-a$. By definition of $\tan$ and $\cot$, we have

$$\cot \left( \frac{\pi }{2}-a\right) =\frac{x^{\prime }}{y^{\prime }},\qquad\tan a=\frac{y}{x}.$$

Since (see picture) $x^{\prime }=y$ and $y^{\prime }=x$, then

$$\cot \left( \frac{\pi }{2}-a\right) =\frac{x^{\prime }}{y^{\prime }}=\frac{y% }{x}=\tan a$$

Added 3: In response to a comment I suggest to use the following triangle to establish the equality between the tangent of an angle and the cotangent of its complement.

$$\cot \alpha=\frac{BC}{AB}=\tan (90{{}^\circ}-\alpha)$$

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I don't really get what is being done but I guess I can just memorize to swap in a negative sign. –  Adam Jun 5 '11 at 20:33
@Adam: The better way is to understand why these identities are valid. My suggestion is that you draw in the trigonometric circle an angle $0\le a\le \pi/2$ and its complement. –  Américo Tavares Jun 5 '11 at 20:40
I tend to think of the underlying "why" as relating to $\theta\mapsto\frac{\pi}{2}-\theta$ as a reflection over $y=x$, which is equivalent to $(x,y)\mapsto(y,x)$, and since $(\cos\theta,\sin\theta)$ is the image of $(1,0)$ under a rotation of $\theta$ centered at the origin, swapping $x$ and $y$ has the effect of swapping sine and cosine... of course, thinking of it that way internally is not the same as thinking it's a useful explanation for someone else... –  Isaac Jun 5 '11 at 20:44
@Adam: I corrected the picture. If it is "far more advanced than anything I [you] have learned so far", I ask how did you learn the trigonometric direct functions cos, sen, tan, sec, csc, cot? And how have you been taught trigonometric equations like the one in the question? –  Américo Tavares Jun 5 '11 at 23:06
@Adam: My last comment, I hope. Draw a right triangle with acute angles $x$ and $y$. You know that $x+y=90{{}^\circ}$, hence $y=90{{}^\circ}-x$. If you use the definitions of $\tan x$ and $\cot y$ what do you get? –  Américo Tavares Jun 5 '11 at 23:17

Given: $\tan \alpha = \cot (\alpha + 10^{\circ}), \quad$ find $\alpha$.

We have that $$\cot a=\tan \left( 90^\circ-a\right)$$ for any acute angle $a$.

In this problem, we have that $a = \alpha + 10$, so substituting for $a$ in the formula above tells us that $$\cot(\alpha + 10)=\tan(90^\circ-(\alpha + 10^\circ)) = \tan(90^\circ - 10^\circ - \alpha) = \tan(80^\circ - \alpha)$$

Back to original problem: $$\tan \alpha = \cot (\alpha + 10^{\circ})$$ $$\tan(\alpha) = \tan(80^\circ - \alpha)$$

Therefore, an easy way to find a solution for $\alpha$ is to notice that the last equation above is true when $$\alpha = 80 - \alpha$$ $$2(\alpha) = 80^\circ$$ $$\alpha = \frac{80^\circ}{2} = 40^\circ$$

$\qquad\qquad\qquad \therefore$ A solution to $\tan \alpha = \cot (\alpha + 10^{\circ})$ is given by $\alpha = 40^\circ$.

Note: this provides one solution, and is not meant to imply that $\alpha = 80 - \alpha$ is the only solution; rather, since the tan(angle1) must be equal to the tangent(angle2), one solution can be found by setting angle1 = angle2.

You might want to bookmark this: Wikipedia List of Trig Identities. It provides a lot of useful identities, and explains why they hold true. For example, you'll find the following useful identities there:

1. $\displaystyle \tan \theta = \frac{\sin \theta}{\cos \theta}$.

2. $\displaystyle \csc \theta = \frac{1}{\sin \theta}$.

3. $\displaystyle \sec \theta = \frac{1}{\cos \theta}$.

4. $\displaystyle \cot \theta = \frac{1}{\tan \theta} = \frac{\cos\theta}{\sin\theta}$.

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The "co" in cofunction is related to the "co" in complement: $$\text{function}(x)=\text{cofunction}(90°-x)$$ A function of an angle is equal to the cofunction of the complement of that angle.

(Alternately, a cofunction of an angle is equal to the function of the complement of that angle: $\text{cofunction}(x)=\text{function}(90°-x)$.)

In your particular equation, look at what happens when you replace $\tan\alpha$ with $\cot(90°-\alpha)$ (which is equal to $\tan\alpha$).

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I don't get it, I have 90-x = cota + cot10 –  Adam Jun 5 '11 at 20:05
@Adam: Your equation was $\tan\alpha=\cot(\alpha +10°)$, so when you replace $\tan\alpha$ with $\cot(90°-\alpha)$, you should have $$\cot(90°-\alpha)=\cot(\alpha +10°).$$ There are many ways for this to be true, but one is easier than the rest: if two expressions are equal, then their cotangents will be equal. What does this suggest about $90°-\alpha$ and $\alpha +10°$? –  Isaac Jun 5 '11 at 20:10
I really don't know I can't even imagine what is going on with this right now, it doesn't make sense to me. I get 80=0? –  Adam Jun 5 '11 at 20:16
@Adam: What did you do to get $80=0$? (In general, it's more useful to talk about what you did to get from one step to the next, rather than what you got at the next step; in particular, I suspect that your thinking may be correct, but you probably know that 80 does not equal 0, so $80=0$ isn't the result we want.) –  Isaac Jun 5 '11 at 20:17
+1 for "The "co" in cofunction is related to the "co" in complement" I didn't know that –  Theta33 Jun 29 '11 at 16:39

Just one small addition to the other answer: Having $\tan\left(\alpha\right) = \tan\left(80^\circ-\alpha\right)$ doesn't mean $\alpha = 80^\circ-\alpha$, only. Note that from graphic of tangent, there's repetition every $\pi$, means $\tan\left(\alpha\right) = \tan\left(\alpha+\pi\right)$.

Well, in your case, this actually doesn't really add any other answer since you have this: "Assume that all angles in which an unknown appears are acute angles."

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