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Maths Notes 11th 12th JEE

Maths Notes 11th 12th JEE

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📈 Análisis del canal de Telegram Maths Notes 11th 12th JEE

El canal Maths Notes 11th 12th JEE (@class_11th_12th_maths_notes_jee) en el segmento lingüístico de Inglés es un actor destacado. Actualmente la comunidad reúne a 22 905 suscriptores, ocupando la posición 8 712 en la categoría Educación y el puesto 19 287 en la región India.

📊 Métricas de audiencia y dinámica

Desde su creación el невідомо, el proyecto ha mostrado un crecimiento acelerado, reuniendo a 22 905 suscriptores.

Según los últimos datos del 09 junio, 2026, el canal mantiene una actividad estable. En los últimos 30 días la variación de miembros fue de -439, y en las últimas 24 horas de -27, conservando un alto alcance.

  • Estado de verificación: No verificado
  • Tasa de interacción (ER): El promedio de interacción de la audiencia es 7.00%. Durante las primeras 24 horas tras publicar, el contenido suele obtener N/A% de reacciones respecto al total de suscriptores.
  • Alcance de las publicaciones: Cada publicación recibe en promedio 1 603 visualizaciones. En el primer día suele acumular 0 visualizaciones.
  • Reacciones e interacción: La audiencia responde de forma activa: el promedio de reacciones por publicación es 6.
  • Intereses temáticos: El contenido se centra en temas clave como biology, chemistry, neet, jee, guidance.

📝 Descripción y política de contenido

El autor describe el recurso como un espacio para expresar opiniones subjetivas:
☎️Contact | Advertising ➭ @ShubhamS90

Gracias a la alta frecuencia de actualizaciones (últimos datos recibidos el 10 junio, 2026), el canal mantiene la vigencia y un amplio alcance. La analítica demuestra que la audiencia interactúa activamente con el contenido, lo que lo convierte en un punto de referencia dentro de la categoría Educación.

22 905
Suscriptores
-2724 horas
-737 días
-43930 días
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➖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
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