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I have seen the following math problem posed online by a high school student (knowing their material, most likely it wasn't given as an exercise):

Find the solutions for the equation $$\tan(\alpha)=\cos(\alpha+33.44^{\circ}).$$

Oddly, the solution he provided was only $\alpha=26.56^\circ$, yet from plotting the equation I see that there are $2$ solutions in each $2\pi$ period. This made me assume that the problem's author had only acute angles in mind (i.e. $0<\alpha<90^\circ$) and that one might be able to solve this geometrically. This is the approach I'm currently trying, though without much luck. Also, the naïve approach of bombarding the equation with trig identities failed miserably.

Ice Tea
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DailyRayleigh
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    Please consider using \cos and \tan to get $\cos$ and $\tan$ instead of simply cos and tan which give $cos$ and $tan$. – Fly by Night Aug 11 '15 at 23:51

1 Answers1

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As far as I can tell, there's nothing particularly special about $33.44^\circ$.

Consider the equation $\tan(\alpha) = \cos(\alpha + \beta)$. With $\cos(\alpha) = c$, $\sin(\alpha) = s$, this becomes

$$ \dfrac{s}{c} = c \cos(\beta) - s \sin(\beta)$$ Eliminating $s$ from this with the help of $c^2 + s^2 = 1$, we get $$ c^4 + 2 c^3 \sin(\beta) + c^2 \cos^2(\beta) - 2 c \sin(\beta) - 1 = 0$$

This quartic equation is irreducible over the field generated by $\sin(\beta)$. So you're not likely to get a very nice algebraic solution in general.

In the case of $\beta = 33.44^\circ$, there are two real solutions for $c$, approximately: $$ c = -.7136633192,\ .8944433266$$ Each corresponds to a value for $s = \sin(\alpha)$: $$ s = .7004888768,\ .4471813227$$ and the values for $\alpha$ are $$ 135.5337601^\circ,\ 26.56298385^\circ$$

Robert Israel
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