202207061152 Solution to 1971-CE-AMATH-I-XX

If $k$ is an integer, simplify

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}$ .

Roughwork.

First,

$\begin{aligned} & \quad \cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg] \\ & = \cos\bigg( 2k\pi+\frac{\pi}{2}+\theta \bigg) \tan \bigg( 2k\pi+\frac{3\pi}{2}-\theta \bigg) \\ \dots\textrm{as } &\textrm{are}\cos (\angle ),\tan (\angle )\textrm{ shifted by full periods, \textit{i.e.}, }k\cdot 2\pi\,\dots \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ \end{aligned}$

Second,

recall that

$\begin{aligned} \textrm{(in radians)} & \quad\textrm{(in degrees)} \\ \pi & = 180^\circ \\ \frac{\pi}{2} & = 90^\circ \\ \frac{3\pi}{2} & = 270^\circ\textrm{ \scriptsize{OR} }-90^\circ = -\frac{\pi}{2}\\ \end{aligned}$

Hence,

$\begin{aligned} & \quad \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{3\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(-\frac{\pi}{2}-\theta \bigg) \\ & = \cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg[ -\bigg(\frac{\pi}{2}+\theta \bigg)\bigg] \\ \dots &\textrm{ as }\tan(-\theta )=-\tan\theta \enspace\dots \\ & = -\cos\bigg(\frac{\pi}{2}+\theta\bigg) \tan \bigg(\frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$ .

Recall that

$\begin{aligned} \because \enspace & \tan\theta = \frac{\sin\theta}{\cos\theta} \\ \therefore \enspace & \cos\theta\tan\theta = \sin\theta \\ \end{aligned}$ .

Hence,

$\begin{aligned} &\quad -\cos\bigg( \frac{\pi}{2}+\theta\bigg)\tan\bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\sin\bigg( \frac{\pi}{2}+\theta\bigg) \\ \end{aligned}$

Third,

recall that

$\displaystyle{\sin \bigg(\theta\pm\frac{\pi}{2}\bigg)}=\pm\cos\theta$ .

$\begin{aligned} &\quad -\sin \bigg( \frac{\pi}{2}+\theta\bigg) \\ & = -\cos\theta \\ \end{aligned}$

Answer.

$\displaystyle{\cos\bigg[ \frac{1}{2}(4k+1)\pi +\theta \bigg]\tan\bigg[\frac{1}{2}(4k+3)\pi -\theta \bigg]}=-\cos\theta$ .

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