Maths Notes 11th 12th JEE
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Show more📈 Analytical overview of Telegram channel Maths Notes 11th 12th JEE
Channel Maths Notes 11th 12th JEE (@class_11th_12th_maths_notes_jee) in the English language segment is an active participant. Currently, the community unites 22 905 subscribers, ranking 8 712 in the Education category and 19 287 in the India region.
📊 Audience metrics and dynamics
Since its creation on невідомо, the project has demonstrated rapid growth, gathering an audience of 22 905 subscribers.
According to the latest data from 09 June, 2026, the channel demonstrates stable activity. Although there has been a change in the number of participants by -439 over the last 30 days and by -27 over the last 24 hours, overall reach remains high.
- Verification status: Not verified
- Engagement rate (ER): The average audience engagement rate is 7.00%. Within the first 24 hours after publication, content typically collects N/A% reactions from the total number of subscribers.
- Post reach: On average, each post receives 1 603 views. Within the first day, a publication typically gains 0 views.
- Reactions and interaction: The audience actively supports content: the average number of reactions per post is 6.
- Thematic interests: Content is focused on key topics such as biology, chemistry, neet, jee, guidance.
📝 Description and content policy
The author describes the resource as a platform for expressing subjective opinions:
“☎️Contact | Advertising ➭ @ShubhamS90”
Thanks to the high frequency of updates (latest data received on 10 June, 2026), the channel maintains relevance and a high level of publication reach. Analytics show that the audience actively interacts with content, making it an important point of influence in the Education category.
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| 3 | ➖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 | 2 256 |
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