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1.3 Trigonometriya — Masalalar

Jami: 25 ta | Yechim bilan: ✅

Asosiy Masalalar (1-10)​

Masala 1 ⭐⭐​

Quyidagilarni hisoblang (kalkulyatorsiz): a) sin⁡30°+cos⁡60°\sin 30° + \cos 60° b) tan⁡45°⋅cos⁡0°\tan 45° \cdot \cos 0° c) sin⁡245°+cos⁡245°\sin^2 45° + \cos^2 45°

Yechim

a) sin⁡30°+cos⁡60°=12+12=1\sin 30° + \cos 60° = \frac{1}{2} + \frac{1}{2} = 1

b) tan⁡45°⋅cos⁡0°=1⋅1=1\tan 45° \cdot \cos 0° = 1 \cdot 1 = 1

c) sin⁡245°+cos⁡245°=(22)2+(22)2=12+12=1\sin^2 45° + \cos^2 45° = (\frac{\sqrt{2}}{2})^2 + (\frac{\sqrt{2}}{2})^2 = \frac{1}{2} + \frac{1}{2} = 1

Masala 2 ⭐⭐​

60°60° ni radianga, 3π4\frac{3\pi}{4} ni gradusga o'tkazing.

Yechim

60°=60×π180=π360° = 60 \times \frac{\pi}{180} = \frac{\pi}{3} rad

3π4=3π4×180π=3×1804=135°\frac{3\pi}{4} = \frac{3\pi}{4} \times \frac{180}{\pi} = \frac{3 \times 180}{4} = 135°

Masala 3 ⭐⭐​

sin⁡θ=0.6\sin\theta = 0.6 va θ\theta birinchi chorakda. cos⁡θ\cos\theta va tan⁡θ\tan\theta ni toping.

Yechim

sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 cos⁡2θ=1−0.36=0.64\cos^2\theta = 1 - 0.36 = 0.64 cos⁡θ=0.8\cos\theta = 0.8 (birinchi chorakda ijobiy)

tan⁡θ=sin⁡θcos⁡θ=0.60.8=0.75\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{0.6}{0.8} = 0.75

Masala 4 ⭐⭐​

sin⁡75°\sin 75° ni hisoblang (75°=45°+30°75° = 45° + 30° ishlatib).

Yechim

sin⁡75°=sin⁡(45°+30°)\sin 75° = \sin(45° + 30°) =sin⁡45°cos⁡30°+cos⁡45°sin⁡30°= \sin 45° \cos 30° + \cos 45° \sin 30° =22⋅32+22⋅12= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} =64+24=6+24= \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} ≈0.966\approx 0.966

Masala 5 ⭐⭐​

cos⁡2θ=0.5\cos 2\theta = 0.5 bo'lsa, cos⁡θ\cos\theta ni toping (0°<θ<90°0° < \theta < 90°).

Yechim

cos⁡2θ=2cos⁡2θ−1=0.5\cos 2\theta = 2\cos^2\theta - 1 = 0.5 2cos⁡2θ=1.52\cos^2\theta = 1.5 cos⁡2θ=0.75\cos^2\theta = 0.75 cos⁡θ=0.75=32≈0.866\cos\theta = \sqrt{0.75} = \frac{\sqrt{3}}{2} \approx 0.866

θ=30°\theta = 30°

Masala 6 ⭐⭐​

Uchburchakda a=5a = 5, b=7b = 7, C=60°C = 60°. cc tomonni toping.

Yechim

Kosinus teoremasi: c2=a2+b2−2abcos⁡Cc^2 = a^2 + b^2 - 2ab\cos C c2=25+49−2(5)(7)cos⁡60°c^2 = 25 + 49 - 2(5)(7)\cos 60° c2=74−70⋅0.5=74−35=39c^2 = 74 - 70 \cdot 0.5 = 74 - 35 = 39 c=39≈6.24c = \sqrt{39} \approx 6.24

Masala 7 ⭐⭐​

Uchburchakda a=8a = 8, b=6b = 6, c=10c = 10. CC burchakni toping.

Yechim

cos⁡C=a2+b2−c22ab\cos C = \frac{a^2 + b^2 - c^2}{2ab} =64+36−1002⋅8⋅6=096=0= \frac{64 + 36 - 100}{2 \cdot 8 \cdot 6} = \frac{0}{96} = 0

C=arccos⁡(0)=90°C = \arccos(0) = 90°

Bu to'g'ri burchakli uchburchak (62+82=1026^2 + 8^2 = 10^2).

Masala 8 ⭐⭐​

atan2(−3,4)\text{atan2}(-3, 4) ni hisoblang.

Yechim

x=4>0x = 4 > 0, y=−3<0y = -3 < 0 — IV chorak

θ=arctan⁡(−34)=arctan⁡(−0.75)≈−36.87°≈−0.644\theta = \arctan(\frac{-3}{4}) = \arctan(-0.75) \approx -36.87° \approx -0.644 rad

Masala 9 ⭐⭐​

Robot qo'lining birinchi bo'g'ini L1=1L_1 = 1 m, burchagi θ1=45°\theta_1 = 45°. Oxirgi nuqta koordinatalarini toping.

Yechim

x=L1cos⁡θ1=1⋅cos⁡45°=22≈0.707x = L_1 \cos\theta_1 = 1 \cdot \cos 45° = \frac{\sqrt{2}}{2} \approx 0.707 m

y=L1sin⁡θ1=1⋅sin⁡45°=22≈0.707y = L_1 \sin\theta_1 = 1 \cdot \sin 45° = \frac{\sqrt{2}}{2} \approx 0.707 m

Masala 10 ⭐⭐​

sin⁡θ=513\sin\theta = \frac{5}{13} (θ\theta birinchi chorakda). sin⁡2θ\sin 2\theta ni toping.

Yechim

cos⁡θ=1−sin⁡2θ=1−25169=144169=1213\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13}

sin⁡2θ=2sin⁡θcos⁡θ=2⋅513⋅1213=120169≈0.710\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{5}{13} \cdot \frac{12}{13} = \frac{120}{169} \approx 0.710

O'rtacha Masalalar (11-18)​

Masala 11 ⭐⭐⭐​

Sinus teoremasi bilan: A=30°A = 30°, B=45°B = 45°, a=10a = 10. bb ni toping.

Yechim

asin⁡A=bsin⁡B\frac{a}{\sin A} = \frac{b}{\sin B}

10sin⁡30°=bsin⁡45°\frac{10}{\sin 30°} = \frac{b}{\sin 45°}

100.5=b22\frac{10}{0.5} = \frac{b}{\frac{\sqrt{2}}{2}}

b=20⋅22=102≈14.14b = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2} \approx 14.14

Masala 12 ⭐⭐⭐​

2 bo'g'inli robot: L1=0.5L_1 = 0.5 m, L2=0.3L_2 = 0.3 m. End effector (0.6,0.4)(0.6, 0.4) da bo'lishi uchun θ1\theta_1 va θ2\theta_2 ni toping.

Yechim

r2=x2+y2=0.36+0.16=0.52r^2 = x^2 + y^2 = 0.36 + 0.16 = 0.52 r=0.721r = 0.721 m

cos⁡θ2=r2−L12−L222L1L2=0.52−0.25−0.092(0.5)(0.3)=0.180.3=0.6\cos\theta_2 = \frac{r^2 - L_1^2 - L_2^2}{2L_1L_2} = \frac{0.52 - 0.25 - 0.09}{2(0.5)(0.3)} = \frac{0.18}{0.3} = 0.6

θ2=arccos⁡(0.6)≈53.13°\theta_2 = \arccos(0.6) \approx 53.13°

sin⁡θ2=1−0.36=0.8\sin\theta_2 = \sqrt{1 - 0.36} = 0.8

α=atan2(y,x)=atan2(0.4,0.6)≈33.69°\alpha = \text{atan2}(y, x) = \text{atan2}(0.4, 0.6) \approx 33.69°

β=atan2(L2sin⁡θ2,L1+L2cos⁡θ2)=atan2(0.24,0.68)≈19.44°\beta = \text{atan2}(L_2\sin\theta_2, L_1 + L_2\cos\theta_2) = \text{atan2}(0.24, 0.68) \approx 19.44°

θ1=α−β=33.69°−19.44°≈14.25°\theta_1 = \alpha - \beta = 33.69° - 19.44° \approx 14.25°

Masala 13 ⭐⭐⭐​

Dron ϕ=10°\phi = 10° roll burchagida uchmoqda. Gravitatsiya kuchining gorizontal komponenti qancha? (m=1m = 1 kg, g=10g = 10 m/s²)

Yechim

Gravitatsiya kuchi: Fg=mg=10F_g = mg = 10 N (vertikal pastga)

Roll burchagida gorizontal komponent: Fh=Fgsin⁡ϕ=10sin⁡10°=10×0.174≈1.74F_h = F_g \sin\phi = 10 \sin 10° = 10 \times 0.174 \approx 1.74 N

Masala 14 ⭐⭐⭐​

Aylanish matritsasidan burchakni ajratib oling:

R=[0.866−0.50.50.866]R = \begin{bmatrix} 0.866 & -0.5 \\ 0.5 & 0.866 \end{bmatrix}

Yechim

R=[cos⁡θ−sin⁡θsin⁡θcos⁡θ]R = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}

cos⁡θ=0.866\cos\theta = 0.866, sin⁡θ=0.5\sin\theta = 0.5

θ=atan2(sin⁡θ,cos⁡θ)=atan2(0.5,0.866)=30°\theta = \text{atan2}(\sin\theta, \cos\theta) = \text{atan2}(0.5, 0.866) = 30°

Masala 15 ⭐⭐⭐​

tan⁡θ+cot⁡θ=2\tan\theta + \cot\theta = 2 bo'lsa, sin⁡2θ\sin 2\theta ni toping.

Yechim

tan⁡θ+cot⁡θ=sin⁡θcos⁡θ+cos⁡θsin⁡θ=sin⁡2θ+cos⁡2θsin⁡θcos⁡θ=1sin⁡θcos⁡θ=2\tan\theta + \cot\theta = \frac{\sin\theta}{\cos\theta} + \frac{\cos\theta}{\sin\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin\theta\cos\theta} = \frac{1}{\sin\theta\cos\theta} = 2

sin⁡θcos⁡θ=0.5\sin\theta\cos\theta = 0.5

sin⁡2θ=2sin⁡θcos⁡θ=2×0.5=1\sin 2\theta = 2\sin\theta\cos\theta = 2 \times 0.5 = 1

2θ=90°2\theta = 90°, ya'ni θ=45°\theta = 45°

Masala 16 ⭐⭐⭐​

Uchburchak yuzini toping: a=7a = 7, b=8b = 8, C=120°C = 120°.

Yechim

S=12absin⁡C=12(7)(8)sin⁡120°S = \frac{1}{2}ab\sin C = \frac{1}{2}(7)(8)\sin 120° =28×32=143≈24.25= 28 \times \frac{\sqrt{3}}{2} = 14\sqrt{3} \approx 24.25

Masala 17 ⭐⭐⭐​

Dron A dan B ga uchmoqda. A(0, 0), B(100, 50) m. Yaw burchakni toping.

Yechim

ψ=atan2(Δy,Δx)=atan2(50,100)\psi = \text{atan2}(\Delta y, \Delta x) = \text{atan2}(50, 100) =arctan⁡(0.5)≈26.57°= \arctan(0.5) \approx 26.57°

Masala 18 ⭐⭐⭐​

arcsin⁡(sin⁡200°)\arcsin(\sin 200°) ni hisoblang.

Yechim

sin⁡200°=sin⁡(180°+20°)=−sin⁡20°≈−0.342\sin 200° = \sin(180° + 20°) = -\sin 20° \approx -0.342

arcsin⁡\arcsin ning oralig'i [−90°,90°][-90°, 90°]

arcsin⁡(−0.342)≈−20°\arcsin(-0.342) \approx -20°

Murakkab Masalalar (19-25)​

Masala 19 ⭐⭐⭐⭐​

3D aylanish: Euler burchaklari ϕ=30°\phi = 30°, θ=45°\theta = 45°, ψ=60°\psi = 60° (ZYX). (1,0,0)(1, 0, 0) vektorni aylantiring.

Yechim

R=Rz(ψ)Ry(θ)Rx(ϕ)R = R_z(\psi) R_y(\theta) R_x(\phi)

Rx(30°)=[10000.866−0.500.50.866]R_x(30°) = \begin{bmatrix} 1 & 0 & 0 \\ 0 & 0.866 & -0.5 \\ 0 & 0.5 & 0.866 \end{bmatrix}

Ry(45°)=[0.70700.707010−0.70700.707]R_y(45°) = \begin{bmatrix} 0.707 & 0 & 0.707 \\ 0 & 1 & 0 \\ -0.707 & 0 & 0.707 \end{bmatrix}

Rz(60°)=[0.5−0.86600.8660.50001]R_z(60°) = \begin{bmatrix} 0.5 & -0.866 & 0 \\ 0.866 & 0.5 & 0 \\ 0 & 0 & 1 \end{bmatrix}

R=RzRyRxR = R_z R_y R_x (matritsalarni ko'paytiring)

v′=R⋅[100]≈[0.3540.612−0.707]v' = R \cdot \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \approx \begin{bmatrix} 0.354 \\ 0.612 \\ -0.707 \end{bmatrix}

Masala 20 ⭐⭐⭐⭐​

Gimbal lock: θ=90°\theta = 90° (pitch) da ϕ\phi va ψ\psi ning ta'sirini tahlil qiling.

Yechim

θ=90°\theta = 90° da: sin⁡θ=1\sin\theta = 1, cos⁡θ=0\cos\theta = 0

R=[0sin⁡(ϕ−ψ)cos⁡(ϕ−ψ)0cos⁡(ϕ−ψ)−sin⁡(ϕ−ψ)−100]R = \begin{bmatrix} 0 & \sin(\phi-\psi) & \cos(\phi-\psi) \\ 0 & \cos(\phi-\psi) & -\sin(\phi-\psi) \\ -1 & 0 & 0 \end{bmatrix}

Faqat (ϕ−ψ)(\phi - \psi) farqi muhim — alohida ϕ\phi va ψ\psi ni aniqlash mumkin emas!

Bu gimbal lock — bir daraja erkinlik yo'qoldi.

Masala 21 ⭐⭐⭐⭐​

Radar antennasi maqsadni kuzatmoqda. Azimut α=45°\alpha = 45°, elevatsiya β=30°\beta = 30°. Birlik yo'nalish vektorini toping.

Yechim

Sferik koordinatalar: x=cos⁡βcos⁡α=cos⁡30°cos⁡45°=0.866×0.707≈0.612x = \cos\beta \cos\alpha = \cos 30° \cos 45° = 0.866 \times 0.707 \approx 0.612 y=cos⁡βsin⁡α=cos⁡30°sin⁡45°=0.866×0.707≈0.612y = \cos\beta \sin\alpha = \cos 30° \sin 45° = 0.866 \times 0.707 \approx 0.612 z=sin⁡β=sin⁡30°=0.5z = \sin\beta = \sin 30° = 0.5

d^=(0.612,0.612,0.5)\hat{d} = (0.612, 0.612, 0.5)

Tekshirish: ∣d^∣=0.374+0.374+0.25=1|\hat{d}| = \sqrt{0.374 + 0.374 + 0.25} = 1 ✓

Masala 22 ⭐⭐⭐⭐​

GPS: A(41.0°N, 69.0°E) dan B(39.0°N, 67.0°E) gacha bearing (yo'nalish) ni toping.

Yechim

Δλ=67°−69°=−2°=−0.0349\Delta\lambda = 67° - 69° = -2° = -0.0349 rad ϕ1=41°=0.716\phi_1 = 41° = 0.716 rad ϕ2=39°=0.681\phi_2 = 39° = 0.681 rad

y=sin⁡(Δλ)cos⁡ϕ2=sin⁡(−0.0349)×cos⁡(0.681)y = \sin(\Delta\lambda)\cos\phi_2 = \sin(-0.0349) \times \cos(0.681) =−0.0349×0.777=−0.0271= -0.0349 \times 0.777 = -0.0271

x=cos⁡ϕ1sin⁡ϕ2−sin⁡ϕ1cos⁡ϕ2cos⁡(Δλ)x = \cos\phi_1\sin\phi_2 - \sin\phi_1\cos\phi_2\cos(\Delta\lambda) =0.755×0.629−0.656×0.777×0.9994= 0.755 \times 0.629 - 0.656 \times 0.777 \times 0.9994 =0.475−0.509=−0.034= 0.475 - 0.509 = -0.034

bearing=atan2(−0.0271,−0.034)≈−141.5°+360°=218.5°\text{bearing} = \text{atan2}(-0.0271, -0.034) \approx -141.5° + 360° = 218.5°

(Janubi-g'arbga)

Masala 23 ⭐⭐⭐⭐​

Pendulum harakati: θ(t)=0.2cos⁡(2t)\theta(t) = 0.2\cos(2t) rad. t=π/4t = \pi/4 s da burchak tezligini toping.

Yechim

ω=dθdt=−0.2×2sin⁡(2t)=−0.4sin⁡(2t)\omega = \frac{d\theta}{dt} = -0.2 \times 2 \sin(2t) = -0.4\sin(2t)

t=π/4t = \pi/4 da: ω=−0.4sin⁡(π/2)=−0.4×1=−0.4\omega = -0.4\sin(\pi/2) = -0.4 \times 1 = -0.4 rad/s

Masala 24 ⭐⭐⭐⭐​

Quadcopter propeller ω=500\omega = 500 rad/s aylanmoqda. 0.1 s da necha aylanish?

Yechim

Burchak o'zgarishi: Δθ=ω×Δt=500×0.1=50\Delta\theta = \omega \times \Delta t = 500 \times 0.1 = 50 rad

Aylanishlar soni: n=Δθ2π=502π≈7.96n = \frac{\Delta\theta}{2\pi} = \frac{50}{2\pi} \approx 7.96 aylanish

Masala 25 ⭐⭐⭐⭐​

Uchta sensor o'rnatlgan: θ1=0°\theta_1 = 0°, θ2=120°\theta_2 = 120°, θ3=240°\theta_3 = 240°. Har birining birlik vektorini va ularning yig'indisini toping.

Yechim

v⃗1=(cos⁡0°,sin⁡0°)=(1,0)\vec{v}_1 = (\cos 0°, \sin 0°) = (1, 0) v⃗2=(cos⁡120°,sin⁡120°)=(−0.5,0.866)\vec{v}_2 = (\cos 120°, \sin 120°) = (-0.5, 0.866) v⃗3=(cos⁡240°,sin⁡240°)=(−0.5,−0.866)\vec{v}_3 = (\cos 240°, \sin 240°) = (-0.5, -0.866)

Yig'indi: v⃗1+v⃗2+v⃗3=(1−0.5−0.5,0+0.866−0.866)=(0,0)\vec{v}_1 + \vec{v}_2 + \vec{v}_3 = (1 - 0.5 - 0.5, 0 + 0.866 - 0.866) = (0, 0)

Teng taqsimlangan sensorlar yig'indisi nolga teng!

✅ Tekshirish Ro'yxati​

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  • 11-18: O'rtacha masalalar
  • 19-25: Murakkab masalalar

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