Page:Calculus Made Easy.pdf/186

supposed to increase by the addition to it of the small angle $d\theta$–an element of angle–the height of $y$, the sine, will be increased by a small element $dy$. The new height $y+dy$ will be the sine of the new angle $\theta +d\theta$, or, stating it as an equation,

$y+dy=\sin(\theta +d\theta )$;

and subtracting from this the first equation gives

$dy=\sin(\theta +d\theta )-\sin \theta$.

The quantity on the right-hand side is the difference between two sines, and books on trigonometry tell us how to work this out. For they tell us that if $M$ and $N$ are two different angles,

$\sin M-\sin N=2\cos {\frac {M+N}{2}}\cdot \sin {\frac {M-N}{2}}$.

If, then, we put $M=\theta +d\theta$ for one angle, and $N=\theta$ for the other, we may write

${\begin{aligned}dy&=2\cos {\frac {\theta +d\theta +\theta }{2}}\cdot \sin {\frac {\theta +d\theta -\theta }{2}},\\{\text{or,}}\quad \quad \quad dy&=2\cos(\theta +{\tfrac {1}{2}}d\theta )\cdot \sin {\tfrac {1}{2}}d\theta .\end{aligned}}$

But if we regard $d\theta$ as indefinitely small, then in the limit we may neglect ${\frac {1}{2}}d\theta$ by comparison with $\theta$, and may also take $\sin {\frac {1}{2}}d\theta$ as being the same as ${\frac {1}{2}}d\theta$. The equation then becomes:

${\begin{aligned}dy&=2\cos \theta \times {\tfrac {1}{2}}d\theta ;\\dy&=\cos \theta \cdot d\theta ,\\{\text{and, finally,}}\quad \quad \quad \quad \quad {\dfrac {dy}{d\theta }}&=\cos \theta .\end{aligned}}$