Maths Notes 11th 12th JEE
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Maths Notes 11th 12th JEE (@class_11th_12th_maths_notes_jee) Ingliz til segmentidagi kanali faol ishtirokchi. Hozirda hamjamiyat 21 926 obunachidan iborat bo'lib, Taʼlim toifasida 9 170-o'rinni va Hindiston mintaqasida 18 557-o'rinni egallagan.
📊 Auditoriya ko‘rsatkichlari va dinamika
невідомо sanasidan buyon loyiha tez o‘sib, 21 926 obunachiga ega bo‘ldi.
27 Oktabr, 2026 dagi oxirgi ma’lumotlarga ko‘ra kanal barqaror faollikka ega. Oxirgi 30 kunda obunachilar soni -283 ga, so‘nggi 24 soatda esa -5 ga o‘zgardi va umumiy qamrov yuqori darajada qolmoqda.
- Tasdiqlash holati: Tasdiqlanmagan
- Jalb etish (ER): Auditoriya o‘rtacha 3.63% darajada jalb etiladi. Nashrdan keyingi dastlabki 24 soatda kontent odatda umumiy obunachilar sonining 8.47% ini tashkil etuvchi reaksiyalarni to‘playdi.
- Post qamrovi: Har bir post o‘rtacha 797 marta ko‘riladi; birinchi sutkada odatda 1 857 ta ko‘rish yig‘iladi.
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Yuqori yangilanish chastotasi (oxirgi ma’lumot 28 Oktabr, 2026 da olingan) sababli kanal doimo dolzarb va katta qamrovli bo‘lib qoladi. Analitika auditoriya kontent bilan faol hamkorlik qilishini, uni Taʼlim toifasidagi muhim ta’sir nuqtasiga aylantirishini ko‘rsatadi.
21 926
Obunachilar
-524 soatlar
-477 kun
-28330 kun
Ma'lumotlar yuklanmoqda
O'xshash kanallar
Taglar buluti
Kirish va chiqish esdaliklari
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| 10 | ➖Double or Triple -Angle Identities
1) sin 2x = 2sin x cos x
2) cos2x = cos2x – sin2x = 1 – 2sin2x = 2cos2x – 1
3) tan 2x = 2 tan x / (1-tan 2x)
4) sin 3x = 3 sin x – 4 sin3x
5) cos3x = 4 cos3x – 3 cosx
6) tan 3x = (3 tan x - tan3x) / (1- 3tan 2x)
➖For angles A, B and C, we have
1) sin (A + B +C) = sinAcosBcosC + cosAsinBcosC + cosAcosBsinC - sinAsinBsinC
2) cos (A + B +C) = cosAcosBcosC- cosAsinBsinC - sinAcosBsinC - sinAsinBcosC
3) tan (A + B +C) = [tan A + tan B + tan C –tan A tan B tan C]/ [1- tan Atan B - tan B tan C –tan A tan C
4) cot (A + B +C) = [cot A cot B cot C – cotA - cot B - cot C]/ [cot A cot B + cot Bcot C + cot A cotC–1]
➖List of some other trigonometric formulas:
1) 2sinAcosB = sin(A + B) + sin (A - B)
2) 2cosAsinB = sin(A + B) - sin (A - B)
3) 2cosAcosB = cos(A + B) + cos(A - B)
4) 2sinAsinB = cos(A - B) - cos (A + B)
5) sin A + sin B = 2 sin [(A+B)/2] cos [(A-B)/2]
6) sin A - sin B = 2 sin [(A-B)/2] cos [(A+B)/2]
7) cosA + cos B = 2 cos [(A+B)/2] cos [(A-B)/2]
8) cosA - cos B = 2 sin [(A+B)/2] sin [(B-A)/2]
9) tanA ± tanB = sin (A ± B)/ cos A cos B
10)cot A ± cot B = sin (B ± A)/ sin A sin B
➖Method of solving a trigonometric equation:
1) If possible, reduce the equation in terms of any one variable, preferably x. Then solve the equation as you used to in case of a single variable.
2) Try to derive the linear/algebraic simultaneous equations from the given trigonometric equations and solve them as algebraic simultaneous equations.
3) At times, you might be required to make certain substitutions. It would be beneficial when the system has only two trigonometric functions.
➖Some results which are useful for solving trigonometric equations:
1) sin θ = sina and cosθ = cosa ⇒ θ = 2nπ + a
2) sin θ = 0 ⇒ θ = nπ
3) cosθ = 0 ⇒ θ = (2n + 1)π/2
4) tan θ = 0 ⇒ θ = nπ
5) sinθ = sina⇒ θ = nπ + (-1)na where a ∈ [–π/2, π/2]
6) cosθ= cos a ⇒ θ = 2nπ ± a, where a ∈[0,π]
7) tanθ = tana⇒ θ = nπ+ a, where a ∈[–π/2, π/2]
8) sinθ = 1 ⇒ θ= (4n + 1)π/2
9) sin θ = -1 ⇒ θ = (4n - 1) π /2
10) sin θ = -1 ⇒ θ = (2n +1) π /2
11) |sinθ| = 1⇒ θ =2nπ
12) cosθ = 1 ⇒ θ =(2n + 1)
13) |cosθ| = 1⇒ θ =nπ | 3 576 |
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