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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).
[Added later] I also need to solve: $\sec(3 \beta + 10) = \csc( \beta + 8)$ (angles are in degrees). |
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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 $\alpha =40{{}^\circ}$. $$\begin{eqnarray*}\tan \alpha &=&\cot \left( \alpha +10{{}^\circ}\right) =\tan \left( 90{{}^\circ}-\alpha -10{{}^\circ}\right) \\&& \\&\Rightarrow &\alpha =90{{}^\circ}-\alpha -10{{}^\circ}\Leftrightarrow \alpha =40{{}^\circ} \end{eqnarray*}$$ Added: For your information the other identities for complementary angles are: $$\begin{eqnarray*} \sin \left( \frac{\pi }{2}-a\right) &=&\cos a \\ \cos \left( \frac{\pi }{2}-a\right) &=&\sin a \\ \tan \left( \frac{\pi }{2}-a\right) &=&\cot a \\ \csc \left( \frac{\pi }{2}-a\right) &=&\sec a \\ \sec \left( \frac{\pi }{2}-a\right) &=&\csc a. \end{eqnarray*}$$ 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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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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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:
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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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