Mavzular

Mavzular / Trigonometriya

Trigonometrik nisbatlar va aniq qiymatlar

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

Asosiy ta'riflar

  • ⭐ To'g'ri burchakli uchburchakda: sin⁡θ=qarama-qarshigipotenuzacos⁡θ=yopishgangipotenuzatan⁡θ=qarama-qarshiyopishgan\sin\theta = \frac{\text{qarama-qarshi}}{\text{gipotenuza}} \qquad \cos\theta = \frac{\text{yopishgan}}{\text{gipotenuza}} \qquad \tan\theta = \frac{\text{qarama-qarshi}}{\text{yopishgan}}
  • sin⁡θ=cos⁡(90°−θ)\sin\theta = \cos(90° - \theta), cos⁡θ=sin⁡(90°−θ)\quad \cos\theta = \sin(90° - \theta)
  • Ko'tarilish burchagi — gorizontal va yuqoridagi yo'nalish orasidagi burchak; tushish burchagi — gorizontal va pastdagi yo'nalish orasidagi burchak. Azimut (bearing) — shimoldan soat strelkasi bo'yicha o'lchanadigan burchak.

Aniq qiymatlar

  • ⭐
θ\theta30°30°45°45°60°60°
sin⁡θ\sin\theta12\frac{1}{2}12\frac{1}{\sqrt{2}}32\frac{\sqrt{3}}{2}
cos⁡θ\cos\theta32\frac{\sqrt{3}}{2}12\frac{1}{\sqrt{2}}12\frac{1}{2}
tan⁡θ\tan\theta13\frac{1}{\sqrt{3}}113\sqrt{3}

Istalgan burchak uchun

  • Birlik aylanada θ\theta burchakka mos nuqta: (cos⁡θ, sin⁡θ)(\cos\theta,\ \sin\theta)
  • ⭐ CAST qoidasi — qaysi chorakda qaysi funksiya musbat:
ChorakMusbat
I (0°0°–90°90°)All — hammasi
II (90°90°–180°180°)Sin
III (180°180°–270°270°)Tan
IV (270°270°–360°360°)Cos
  • sin⁡θ=sin⁡(180°−θ)\sin\theta = \sin(180° - \theta), cos⁡θ=sin⁡(θ+90°)\quad \cos\theta = \sin(\theta + 90°)
  • Davr: sin⁡θ\sin\theta va cos⁡θ\cos\theta — 360°360°, tan⁡θ\tan\theta — 180°180°
  • tan⁡θ\tan\theta θ=90°,270°,…\theta = 90°, 270°, \dots da aniqlanmagan (asimptotalar)

Aniq qiymatlar — radianda va chegaraviy burchaklar

θ\theta00π6\frac{\pi}{6}π4\frac{\pi}{4}π3\frac{\pi}{3}π2\frac{\pi}{2}π\pi
sin⁡θ\sin\theta0012\frac{1}{2}22\frac{\sqrt{2}}{2}32\frac{\sqrt{3}}{2}1100
cos⁡θ\cos\theta1132\frac{\sqrt{3}}{2}22\frac{\sqrt{2}}{2}12\frac{1}{2}00−1-1
tan⁡θ\tan\theta0013\frac{1}{\sqrt{3}}113\sqrt{3}—00

Simmetriya formulalari

sin⁡\sincos⁡\costan⁡\tan
−θ-\theta−sin⁡θ-\sin\thetacos⁡θ\cos\theta−tan⁡θ-\tan\theta
180°−θ180° - \thetasin⁡θ\sin\theta−cos⁡θ-\cos\theta−tan⁡θ-\tan\theta
180°+θ180° + \theta−sin⁡θ-\sin\theta−cos⁡θ-\cos\thetatan⁡θ\tan\theta
360°−θ360° - \theta−sin⁡θ-\sin\thetacos⁡θ\cos\theta−tan⁡θ-\tan\theta
90°−θ90° - \thetacos⁡θ\cos\thetasin⁡θ\sin\theta1tan⁡θ\dfrac{1}{\tan\theta}

🔵 Qo'shimcha burchaklar

θ\thetasin⁡θ\sin\thetacos⁡θ\cos\thetatan⁡θ\tan\theta
15°15°6−24\dfrac{\sqrt{6} - \sqrt{2}}{4}6+24\dfrac{\sqrt{6} + \sqrt{2}}{4}2−32 - \sqrt{3}
75°75°6+24\dfrac{\sqrt{6} + \sqrt{2}}{4}6−24\dfrac{\sqrt{6} - \sqrt{2}}{4}2+32 + \sqrt{3}
18°18°5−14\dfrac{\sqrt{5} - 1}{4}——
36°36°—5+14\dfrac{\sqrt{5} + 1}{4}—