# Foirmlean cur-ris

Nuair a bhios sinn a' cur-ris no a' toirt-air-falbh cheàrnan canaidh sinn ceàrn dà-fhillte ris an toradh. Mar eisimpleir, 's e ceàrn dà-fhillte a th' ann an $$30^\circ + 120^\circ$$. Le àireamhair, obraichidh sinn a-mach:

$\sin (30^\circ + 120^\circ ) = \sin (210^\circ ) = - 0.5$

$\sin (30^\circ ) + \sin (120^\circ ) = 1.366\,(gu\,3\,id.)$

Tha seo a' sealltainn nach eil $$\sin (A + B)$$ co-ionann ri $$\sin A + \sin B$$. An àite sin, faodaidh sinn na co-ionannachdan a leanas a chleachdadh:

$\sin (A + B) = \sin A\cos B + \cos A\sin B$

$\sin (A - B) = \sin A\cos B - \cos A\sin B$

$\cos (A + B) = \cos A\cos B - \sin A\sin B$

$\cos (A - B) = \cos A\cos B + \sin A\sin B$

• Bidh foirmlean airson cur-ris air an sgrìobhadh ann an riochd goirid:
• $\sin (A \pm B) = \sin A\cos B \pm \cos A\sin B$
• $\cos (A \pm B) = \cos A\cos B \pm \sin A\sin B$

Seo mar a bhios sinn a' dol an sàs sa cheist seo sa bheil dà phàirt:

Question

1. Sgrìobh $$75^\circ = 45^\circ + 30^\circ$$ agus obraich a-mach an luach mionaideach aig $$\sin 75^\circ$$

2. Obraich a-mach an luach mionaideach aig $$\cos \left( {\frac{{7\pi }}{{12}}} \right)$$

1. $$\sin 75^\circ = \sin (45 + 30)^\circ$$

A' cleachdadh an fhoirmle $$\sin (A + B)$$

$= \sin 45^\circ \cos 30^\circ + \cos 45^\circ \sin 30^\circ$

A' cleachdadh luachan mionaideach air am bu chòir fios a bhith agad:

$= \frac{1}{{\sqrt 2 }} \times \frac{{\sqrt 3 }}{2} + \frac{1}{{\sqrt 2 }} \times \frac{1}{2}$

$= \frac{{\sqrt 3 }}{{2\sqrt 2 }} + \frac{1}{{2\sqrt 2 }}$

$= \frac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$

2. Bhon a tha $$\frac{{7\pi }}{{12}} = \frac{\pi }{3} + \frac{\pi }{4}$$ tha:

$\cos \left( {\frac{{7\pi }}{{12}}} \right) = \cos \left( {\frac{\pi }{3} + \frac{\pi }{4}} \right)$

A' cleachdadh an fhoirmle airson $$\cos (A + B)$$

$= \cos \frac{\pi }{3}\cos \frac{\pi }{4} - \sin \frac{\pi }{3}\sin \frac{\pi }{4}$

A' cleachdadh luachan mionaideach air am bu chòir fios a bhith agad:

$= \frac{1}{2} \times \frac{1}{{\sqrt 2 }} - \frac{{\sqrt 3 }}{2} \times \frac{1}{{\sqrt 2 }}$

$= \frac{{1 - \sqrt 3 }}{{2\sqrt 2 }}$