Maths Notes 11th 12th JEE

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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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