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Matematika / Trigonometriya

Trigonometrik tenglamalar

⭐ — imtihonda eng ko'p ishlatiladigan formulalar. · 🟢 — masalani tezlashtiradi · 🟡 — ilg'or daraja · 🔵 — chuqur o'rganish (olimpiada darajasi)

Tenglamalarni yechish

  • Bosh qiymatlar — kalkulyator sin⁡−1\sin^{-1}, cos⁡−1\cos^{-1}, tan⁡−1\tan^{-1} uchun beradigan qiymat:
FunksiyaBosh qiymat oralig'i
sin⁡−1x\sin^{-1} x−90°≤θ≤90°-90° \le \theta \le 90°
cos⁡−1x\cos^{-1} x0°≤θ≤180°0° \le \theta \le 180°
tan⁡−1x\tan^{-1} x−90°<θ<90°-90° < \theta < 90°
  • ⭐ Usul: bosh qiymatni topasiz, grafik simmetriyasi orqali ikkinchi qiymatni topasiz (sin⁡\sin uchun 180°−α180° - \alpha, cos⁡\cos uchun 360°−α360° - \alpha, tan⁡\tan uchun 180°+α180° + \alpha), so'ng davrni qo'shib/ayirib oraliqdagi barcha yechimlarni yozasiz. sin⁡\sin va cos⁡\cos aralash bo'lsa, avval 11-ayniyat bilan bitta funksiyaga keltirasiz.

Trigonometrik tenglamalarning barcha yechimlari

α\alpha — kalkulyator bergan bosh qiymat:

TenglamaYechimlar (nn — butun son)
sin⁡θ=k\sin\theta = kθ=α+360°n\theta = \alpha + 360°n yoki θ=180°−α+360°n\theta = 180° - \alpha + 360°n
cos⁡θ=k\cos\theta = kθ=±α+360°n\theta = \pm\alpha + 360°n
tan⁡θ=k\tan\theta = kθ=α+180°n\theta = \alpha + 180°n

sin⁡2θ=k\sin 2\theta = k kabi tenglamada avval 2θ2\theta uchun oraliqni ikki baravar kengaytiring, keyin javoblarni 2 ga bo'ling.

Bosh qiymatlar radianda

−π2≤sin⁡−1x≤π20≤cos⁡−1x≤π−π2<tan⁡−1x<π2-\frac{\pi}{2} \le \sin^{-1}x \le \frac{\pi}{2} \qquad 0 \le \cos^{-1}x \le \pi \qquad -\frac{\pi}{2} < \tan^{-1}x < \frac{\pi}{2}

🟡 Formulalar (nn — butun son, α\alpha — bosh qiymat)

TenglamaUmumiy yechim
sin⁡θ=k\sin\theta = kθ=nπ+(−1)nα\theta = n\pi + (-1)^n\alpha
cos⁡θ=k\cos\theta = kθ=2nπ±α\theta = 2n\pi \pm \alpha
tan⁡θ=k\tan\theta = kθ=nπ+α\theta = n\pi + \alpha

🔵 Teskari trigonometrik ayniyatlar

sin⁡−1x+cos⁡−1x=π2tan⁡−1x+tan⁡−11x=π2    (x>0)\sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \qquad \tan^{-1}x + \tan^{-1}\frac{1}{x} = \frac{\pi}{2} \;\;(x > 0)

Mashq qiling

15 ta savol, vaqt bilan, izohlari bilan.

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