InterviewSolution
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Trigonometric formulae for ""CLASS 11th"" |
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Answer» c Identity of Trigonometric Angles:sin (π/2 – A) = cos A & cos (π/2 – A) = sin ASIN (π/2 + A) = cos A & cos (π/2 + A) = – sin Asin (3π/2 – A) = – cos A & cos (3π/2 – A) = – sin Asin (3π/2 + A) = – cos A & cos (3π/2 + A) = sin Asin (π – A) = sin A & cos (π – A) = – cos Asin (π + A) = – sin A & cos (π + A) = – cos Asin (2π – A) = – sin A & cos (2π – A) = cos Asin (2π + A) = sinA cos (2π + A) = cos ACofunction Identity:sin(90°− X) = cos xcos(90°− x) = sin XTAN(90°− x) = cot xcot(90°− x) = tan xsec(90°− x) = cosec xcosec(90°− x) = sec xSum and Difference Trigonometric FORMULA:sin(x + y) = sin(x)cos(y) + cos(x)sin(y)cos(x + y) = cos(x)cos(y) – sin(x)sin(y)tan(x + y) = (tan x + tan y)/(1 − tan x • tan y)sin(x – y) = sin(x)cos(y) – cos(x)sin(y)cos(x – y) = cos(x)cos(y) + sin(x)sin(y)tan(x − y) = (tan x – tan y)/(1 + tan x • tan y)Double Angle Formula:sin(2X) = 2sin(x) • cos(x) = [2tan x/(1 + tan2x)]cos(2x) = cos2(x) – sin2(x) = [(1 - tan2x)/(1 + tan2x)]cos(2x) = 2cos2(x) − 1 = 1 – 2sin2(x)tan(2x) = [2tan(x)]/ [1−tan2(x)]sec (2x) = sec2 x/(2 - sec2 x)csc (2x) = (sec x. csc x)/2Inverse Trigonometric Function:sin-1 (–x) = – sin-1 xcos-1 (–x) = π – cos-1 xtan-1 (–x) = – tan-1 xcosec-1 (–x) = – cosec-1 xsec-1 (–x) = π – sec-1 xcot-1 (–x) = π – cot-1 x |
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