Réitigh sinalpha=?

$$\sin \alpha = ?$$

Difreálaigh w.r.t. α

$$\cos(\alpha )$$

Céimeanna ag Baint Úsáid as Sainiú Díorthaigh

Do fheidhm f\left(x\right), is ionann an díorthach agus teorainn \frac{f\left(x+h\right)-f\left(x\right)}{h} toisc go dtéann h go 0, más ann don teorainn sin.

$$\frac{\mathrm{d}}{\mathrm{d}\alpha }(\sin(\alpha ))=\left(\lim_{h\to 0}\frac{\sin(\alpha +h)-\sin(\alpha )}{h}\right)$$

Úsáid an Fhoirmle Shuime don Síneas.

$$\lim_{h\to 0}\frac{\sin(h+\alpha )-\sin(\alpha )}{h}$$

Fág \sin(\alpha ) as an áireamh.

$$\lim_{h\to 0}\frac{\sin(\alpha )\left(\cos(h)-1\right)+\cos(\alpha )\sin(h)}{h}$$

Athscríobh an teorainn.

$$\left(\lim_{h\to 0}\sin(\alpha )\right)\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\left(\lim_{h\to 0}\cos(\alpha )\right)\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

Bain leas as an rud é go bhfuil \alpha ina thairiseach agus teorainneacha á ríomh agus h ag dul go 0.

$$\sin(\alpha )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\alpha )\left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)$$

Is ionann teorainn \lim_{\alpha \to 0}\frac{\sin(\alpha )}{\alpha } agus 1.

$$\sin(\alpha )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\alpha )$$

Chun an teorainn \lim_{h\to 0}\frac{\cos(h)-1}{h} a luacháil, méadaigh an t-uimhreoir agus an t-ainmneoir faoi \cos(h)+1 ar dtús.

$$\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)=\left(\lim_{h\to 0}\frac{\left(\cos(h)-1\right)\left(\cos(h)+1\right)}{h\left(\cos(h)+1\right)}\right)$$

Méadaigh \cos(h)+1 faoi \cos(h)-1.

$$\lim_{h\to 0}\frac{\left(\cos(h)\right)^{2}-1}{h\left(\cos(h)+1\right)}$$

Baint Úsáid as Aitheantas Píotagarásach.

$$\lim_{h\to 0}-\frac{\left(\sin(h)\right)^{2}}{h\left(\cos(h)+1\right)}$$

Athscríobh an teorainn.

$$\left(\lim_{h\to 0}-\frac{\sin(h)}{h}\right)\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

Is ionann teorainn \lim_{\alpha \to 0}\frac{\sin(\alpha )}{\alpha } agus 1.

$$-\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)$$

Bain leas as an rud go bhfuil \frac{\sin(h)}{\cos(h)+1} leanúnach ag 0.

$$\left(\lim_{h\to 0}\frac{\sin(h)}{\cos(h)+1}\right)=0$$

Ionadaigh an luach 0 isteach sa slonn \sin(\alpha )\left(\lim_{h\to 0}\frac{\cos(h)-1}{h}\right)+\cos(\alpha ).

$$\cos(\alpha )$$

Luacháil

$$\sin(\alpha )$$