Co-aontaran triantanachd le foirmle tuinn
Eisimpleir
Fuasgail \(\sqrt 3 \cos x + \sin x = \sqrt 2\), airson \(0 \le x \le 2\pi\).
Fuasgladh
An toiseach feumaidh sinn \(\sqrt 3 \cos x + \sin x\) a chur dhan riochd \(k\cos (x - \alpha)\).
\(k\cos (x-\alpha) = k\cos x\cos \alpha + k\sin x\sin \alpha\)
\(k\cos \alpha = \sqrt 3\)
\(k\sin \alpha = 1\)
Obraich a-mach \(k\) a' cleachdadh luachan airson nan co-èifeachdan:
\(k = \sqrt {\left( {\sqrt {{3^2}} } \right) + {1^2}}\)
\(= \sqrt {3 + 1}\)
\(= \sqrt 4\)
\(= 2\)
Bhon a tha \(\tan \theta = \frac{{sin\theta }}{{\cos \theta }}\)
\(\tan \alpha = \frac{1}{{\sqrt 3 }}\)
\(\alpha = \frac{\pi }{6}\)
Mar sin \(\sqrt 3 \cos x + \sin x = 2\cos \left( {x - \frac{\pi }{6}} \right)\)
\(2\cos \left( {x - \frac{\pi }{6}} \right) = \sqrt 2\)
\(\cos \left( {x - \frac{\pi }{6}} \right) = \frac{{\sqrt 2 }}{2}\)
\(\cos \left( {x - \frac{\pi }{6}} \right) = \frac{1}{{\sqrt 2 }}\)
Bhon a tha cos dearbhte, tha sinn sa 1d agus sa 4mh cairteal.
A' chiad chairteal
\(x - \frac{\pi }{6} = {\cos ^{ - 1}}\frac{1}{{\sqrt 2 }}\)
\(x - \frac{\pi }{6} = \frac{\pi }{4}\)
\(x = \frac{{5\pi }}{{12}}\)
An ceathramh cairteal
\(x - \frac{\pi }{6} = 2\pi - \frac{\pi }{4}\)
\(x - \frac{\pi }{6} = \frac{{7\pi }}{4}\)
\(x = \frac{{23\pi }}{{12}}\)
Mar sin \(x = \frac{{5\pi }}{{12}},\,\frac{{23\pi }}{{12}}\)
