Math, asked by heenasharif51, 4 months ago

formula of Trigonometry​

Answers

Answered by raushankumar3604
1

Answer:

\huge \underline\bold\red{Answer!!}

By using a right-angled triangle as a reference, the trigonometric functions and identities are derived:

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

The Reciprocal Identities are given as:

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

All these are taken from a right angled triangle. When height and base side of the right triangle are known, we can find out the sine, cosine, tangent, secant, cosecant, and cotangent values using trigonometric formulas. The reciprocal trigonometric identities are also derived by using the trigonometric functions.

Periodicity Identities (in Radians)

These formulas are used to shift the angles by π/2, π, 2π, etc. They are also called co-function identities.

sin (π/2 – A) = cos A & cos (π/2 – A) = sin A

sin (π/2 + A) = cos A & cos (π/2 + A) = – sin A

sin (3π/2 – A) = – cos A & cos (3π/2 – A) = – sin A

sin (3π/2 + A) = – cos A & cos (3π/2 + A) = sin A

sin (π – A) = sin A & cos (π – A) = – cos A

sin (π + A) = – sin A & cos (π + A) = – cos A

sin (2π – A) = – sin A & cos (2π – A) = cos A

sin (2π + A) = sin A & cos (2π + A) = cos A

Co-function Identities (in Degrees)

The co-function or periodic identities can also be represented in degrees as:

sin(90°−x) = cos x

cos(90°−x) = sin x

tan(90°−x) = cot x

cot(90°−x) = tan x

sec(90°−x) = csc x

csc(90°−x) = sec x

Sum & Difference Identities

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 Identities

sin(2x) = 2sin(x) • cos(x) = [2tan x/(1+tan2 x)]

cos(2x) = cos2(x)–sin2(x) = [(1-tan2 x)/(1+tan2 x)]

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)/2

Triple Angle Identities

Sin 3x = 3sin x – 4sin3x

Cos 3x = 4cos3x-3cos x

Tan 3x = [3tanx-tan3x]/[1-3tan2x]

Half Angle Identities

sinx2=±1−cosx2−−−−−−√

cosx2=±1+cosx2−−−−−−√

tan(x2)=1−cos(x)1+cos(x)−−−−−−√

Also, tan(x2)=1−cos(x)1+cos(x)−−−−−−√=(1−cos(x))(1−cos(x))(1+cos(x))(1−cos(x))−−−−−−−−−−−−−√=(1−cos(x))21−cos2(x)−−−−−−−−√=(1−cos(x))2sin2(x)−−−−−−−−√=1−cos(x)sin(x) So, tan(x2)=1−cos(x)sin(x)

Product identities

sinx⋅cosy=sin(x+y)+sin(x−y)2

cosx⋅cosy=cos(x+y)+cos(x−y)2

sinx⋅siny=cos(x−y)−cos(x+y)2

Sum to Product Identities

sinx+siny=2sinx+y2cosx−y2

sinx−siny=2cosx+y2sinx−y2

cosx+cosy=2cosx+y2cosx−y2

cosx−cosy=−2sinx+y2sinx−y2

Inverse Trigonometry Formulas

sin-1 (–x) = – sin-1 x

cos-1 (–x) = π – cos-1 x

tan-1 (–x) = – tan-1 x

cosec-1 (–x) = – cosec-1 x

sec-1 (–x) = π – sec-1 x

cot-1 (–x) = π – cot-1 x

Answered by rajkhan802212
4

Answer:

ohhhh.... I'm read in class 10th from CBSE board.

Are u active in any other Social media sites?

Similar questions