Mavzular

Mavzular / Trigonometriya

Trigonometrik ayniyatlar

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

Asosiy ayniyatlar

  • ⭐ tan⁡θ≡sin⁡θcos⁡θ\tan\theta \equiv \dfrac{\sin\theta}{\cos\theta}
  • ⭐ sin⁡2θ+cos⁡2θ≡1\sin^2\theta + \cos^2\theta \equiv 1, bundan   sin⁡2θ≡1−cos⁡2θ\;\sin^2\theta \equiv 1 - \cos^2\theta,   cos⁡2θ≡1−sin⁡2θ\;\cos^2\theta \equiv 1 - \sin^2\theta
  • Ayniyatni isbotlash usullari: (a) murakkabroq tomonni soddalashtirib, ikkinchi tomonga keltirish; (b) chap−o’ng≡0\text{chap} - \text{o'ng} \equiv 0 ekanini ko'rsatish.

🟡 Boshqa trigonometrik funksiyalar

sec⁡θ=1cos⁡θcosec⁡θ=1sin⁡θcot⁡θ=1tan⁡θ=cos⁡θsin⁡θ\sec\theta = \frac{1}{\cos\theta} \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \cot\theta = \frac{1}{\tan\theta} = \frac{\cos\theta}{\sin\theta}

1+tan⁡2θ=sec⁡2θ1+cot⁡2θ=cosec⁡2θ1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \operatorname{cosec}^2\theta

🟡 Qo'shish formulalari

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B

cos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡B\cos(A \pm B) = \cos A\cos B \mp \sin A\sin B

tan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

🟡 Ikkilangan burchak

sin⁡2A=2sin⁡Acos⁡Atan⁡2A=2tan⁡A1−tan⁡2A\sin 2A = 2\sin A\cos A \qquad \tan 2A = \frac{2\tan A}{1 - \tan^2 A}

cos⁡2A=cos⁡2A−sin⁡2A=1−2sin⁡2A=2cos⁡2A−1\cos 2A = \cos^2 A - \sin^2 A = 1 - 2\sin^2 A = 2\cos^2 A - 1

🟡 Darajani pasaytirish

sin⁡2A=1−cos⁡2A2cos⁡2A=1+cos⁡2A2\sin^2 A = \frac{1 - \cos 2A}{2} \qquad \cos^2 A = \frac{1 + \cos 2A}{2}

Bular ∫sin⁡2x dx\int \sin^2 x\,dx kabi integrallarda kerak bo'ladi.

🟡 RR-shakl (harmonik shakl)

asin⁡θ+bcos⁡θ=Rsin⁡(θ+α)acos⁡θ+bsin⁡θ=Rcos⁡(θ−α)a\sin\theta + b\cos\theta = R\sin(\theta + \alpha) \qquad a\cos\theta + b\sin\theta = R\cos(\theta - \alpha)

R=a2+b2,tan⁡α=baR = \sqrt{a^2 + b^2}, \qquad \tan\alpha = \frac{b}{a}

  • Eng katta qiymati RR, eng kichigi −R-R.

🔵 Uchlangan burchak

sin⁡3A=3sin⁡A−4sin⁡3Acos⁡3A=4cos⁡3A−3cos⁡A\sin 3A = 3\sin A - 4\sin^3 A \qquad \cos 3A = 4\cos^3 A - 3\cos A

🔵 Yig'indini ko'paytmaga aylantirish

sin⁡A+sin⁡B=2sin⁡A+B2cos⁡A−B2sin⁡A−sin⁡B=2cos⁡A+B2sin⁡A−B2\sin A + \sin B = 2\sin\frac{A + B}{2}\cos\frac{A - B}{2} \qquad \sin A - \sin B = 2\cos\frac{A + B}{2}\sin\frac{A - B}{2}

cos⁡A+cos⁡B=2cos⁡A+B2cos⁡A−B2cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A + \cos B = 2\cos\frac{A + B}{2}\cos\frac{A - B}{2} \qquad \cos A - \cos B = -2\sin\frac{A + B}{2}\sin\frac{A - B}{2}

🔵 Ko'paytmani yig'indiga aylantirish

2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)2\sin A\cos B = \sin(A + B) + \sin(A - B)

2cos⁡Acos⁡B=cos⁡(A+B)+cos⁡(A−B)2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\cos A\cos B = \cos(A + B) + \cos(A - B) \qquad 2\sin A\sin B = \cos(A - B) - \cos(A + B)

🔵 tt-formulalar (t=tan⁡θ2t = \tan\frac{\theta}{2})

sin⁡θ=2t1+t2cos⁡θ=1−t21+t2tan⁡θ=2t1−t2\sin\theta = \frac{2t}{1 + t^2} \qquad \cos\theta = \frac{1 - t^2}{1 + t^2} \qquad \tan\theta = \frac{2t}{1 - t^2}

🟡 (θ\theta radianda, kichik)

sin⁡θ≈θtan⁡θ≈θcos⁡θ≈1−θ22\sin\theta \approx \theta \qquad \tan\theta \approx \theta \qquad \cos\theta \approx 1 - \frac{\theta^2}{2}