Physics, asked by vishalsingh1234, 1 year ago

The approximate value of x
where x = Sin2°cos2° , is ?

Answers

Answered by PoojaBurra
10

Given :

The relation X =Sin2°Cos2°

To Find :

The value of X

Solution :

  • We know the trigonometric relation

           Sin2θ = 2SinθCosθ

           SinθCosθ = Sin2θ/2

  • By substituting the value θ=2° in the above relation

           Sin2°Cos2° = Sin2(2)/2

                               = Sin4° / 2

                               =0.06976\2

           Sin2°Cos2° =0.03488

The value of  X =Sin2°Cos2° is 0.03488

Answered by amitnrw
19

Given : x  = sin2°Cos2°

To find : The approximate value of x

Solution:

x  = sin2°Cos2°

=> x  = 2sin2°Cos2°/2

=> x = Sin4°/2

180° = π

=> 1° = π/180

=> 4° = 4π/180

=> 4° = 2π/90

=> x = Sin( 2π/90) / 2

Multiplying & dividing by π/90

=> x = ( π/90) Sin( 2π/90) / (2(π/90) )

=> x = ( π/90)   Sin( 2π/90) / (2π/90 )

2π/90 tends to 0

=> Sin Sin( 2π/90) / (2π/90 ) = 1

=> x =  π/90

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