Page:Calculus Made Easy.pdf/188

Take next the cosine.

Let $y=\cos \theta$.

Now $\cos \theta =\sin \left({\dfrac {\pi }{2}}-\theta \right)$.

Therefore

${\begin{aligned}&{\begin{aligned}dy=d\left(\sin \left({\frac {\pi }{2}}-\theta \right)\right)&=\cos \left({\frac {\pi }{2}}-\theta \right)\times d(-\theta ),\\&=\cos \left({\frac {\pi }{2}}-\theta \right)\times (-d\theta ),\end{aligned}}\\&{\frac {dy}{d\theta }}=-\cos \left({\frac {\pi }{2}}-\theta \right).\end{aligned}}$

And it follows that

${\frac {dy}{d\theta }}=-\sin \theta$.

 

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Lastly, take the tangent. Let

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikisource.org/v1/":): {\displaystyle \begin{align} y &= \tan \theta, \\ dy &= \tan(\theta + d\theta) - \tan\theta. \\ \end{align}}

Expanding, as shown in books on trigonometry,

${\begin{aligned}\tan(\theta +d\theta )&={\frac {\tan \theta +\tan d\theta }{1-\tan \theta \cdot \tan d\theta }};\\{\text{whence}}\quad \quad \quad dy&={\frac {\tan \theta +\tan d\theta }{1-\tan \theta \cdot \tan d\theta }}-\tan \theta \\&={\frac {(1+\tan ^{2}\theta )\tan d\theta }{1-\tan \theta \cdot \tan d\theta }}.\end{aligned}}$