Fuincsean tuinn
acosx + bsinx
Faodaidh sinn abairt sam bith san riochd \(a\cos x + b\sin x\), a sgrìobhadh ann am fear sam bith de na riochdan a leanas:
- \(k\cos (x - \alpha )\)
- \(k\cos (x + \alpha )\)
- \(k\sin (x - \alpha )\)
- \(k\sin (x + \alpha )\)
'S dòcha gun inns a' cheist dhut dè an riochd a chleachdas tu, ach mura h-inns, nì riochd sam bith a' chùis. Mura bi thu cinnteach, 's àbhaist gun obraich \(k\cos (x - \alpha )\).
Tha eisimpleirean air an obrachadh a-mach an seo. Chì thu na pròiseasan, a dh'fheumas tu a leantainn, gus abairt san riochd seo ath-sgrìobhadh.
Question
Sgrìobh \(2\sin x^\circ + 5\cos x^\circ\) san riochd \(k\sin (x + \alpha )^\circ\) far a bheil \(k\textgreater0\) agus \(0^\circ \le \alpha ^\circ \textless 360^\circ\).
Fhuair thu an riochd anns am feum thu an co-aontar a sgrìobhadh. Feumaidh tu a-nis an dà abairt a dhèanamh co-ionann.
\(2\sin x^\circ + 5\cos x^\circ = k\sin (x + \alpha )^\circ\)
Meudaich na camagan air an taobh dheas dhen cho-aontar.
\(= k\sin x^\circ \cos \alpha ^\circ + k\cos x^\circ \sin \alpha ^\circ\)
Dèan na co-èifeachdairean co-ionann:
\(k\cos \alpha ^\circ = 2\)
\(k\sin \alpha ^\circ = 5\)
Obraich a-mach \(k\) a' cleachdadh nan luachan aig na co-èifeachdairean:
\(k = \sqrt {{a^2} + {b^2}}\)
\(= \sqrt {4 + 25}\)
\(= \sqrt {29}\)
Obraich a-mach \(\alpha\)
\(\tan \alpha ^\circ =\frac{{k\sin \alpha ^\circ }}{{k\cos \alpha ^\circ }} = \frac{5}{2}\)
\(\alpha = {\tan ^{ - 1}}\left( {\frac{5}{2}} \right)\)
Tha fios againn gu bheil \(\alpha\) sa chiad chairteal mar \(k\cos \alpha ^\circ \textgreater 0\) agus \(k\sin \alpha ^\circ \textgreater 0\)
\(= 68.2^\circ\)
Sgrìobh a-nis an co-aontar san riochd a chaidh iarraidh ort:
\(2\sin x^\circ + 5\cos x^\circ = \sqrt {29} \sin (x + 68.2)^\circ\)
Question
Sgrìobh \(\cos 2x - \sqrt 3 \sin 2x\) san riochd \(k\cos (2x + \alpha )\) far a bheil \(k\textgreater0\) agus \(0 \le \alpha \le 2\pi\)
\(\cos 2x - \sqrt 3 \sin 2x = k\cos (2x + \alpha )\)
\(= k\cos 2x\cos \alpha - k\sin 2x\sin \alpha\)
\(= k\cos \alpha \cos 2x - k\sin \alpha \sin 2x\)
\(k\cos \alpha = 1\)
'S e \(k\cos \alpha\) an co-èifeachdair aig an teirm \(\cos 2x\)
\(k\sin \alpha = \sqrt 3\)
'S e \(k\sin \alpha\) an co-èifeachdair aig an teirm \(\sin 2x\)
\(k = \sqrt {{1^2} + {{\left( {\sqrt 3 } \right)}^2}}\)
\(= \sqrt {1 + 3}\)
\(= \sqrt 4\)
\(= 2\)
Tha fios againn gu bheil \(\alpha\) sa chiad chairteal mar \(k\cos \alpha \textgreater0\) agus \(k\sin \alpha \textgreater0\).
\(a = {\tan ^{ - 1}}\sqrt 3\)
\(= \frac{\pi }{3}\)
Mar sin:
\(\cos 2x - \sqrt 3 \sin 2x = 2\cos \left( {2x + \frac{\pi }{3}} \right)\)
