y = y 1 +y 2 = a [sin 2π (t/T – x/λ) + sin 2π (t/T + x/λ)] = a [2sin (2πt/T) cos (2πx/λ)] So, y = 2a cos (2πx/λ) sin (2πt/T) how was the second step in terms of COS arrived?

Question

y = y 1 +y 2 = a [sin 2π (t/T – x/λ) + sin 2π (t/T + x/λ)] = a [2sin (2πt/T) cos (2πx/λ)] So, y = 2a cos (2πx/λ) sin (2πt/T) how was the second step in terms of COS arrived?

Vikas TU · Answer

Dear student It is derived by using the formula SinA + Sin B = 2 Sin(A+B)/2 * Cos(A-B)/2 Hope this will help Good Luck Cheers

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y = y 1 +y 2 = a [sin 2π (t/T – x/λ) + sin 2π (t/T + x/λ)] = a [2sin (2πt/T) cos (2πx/λ)] So, y = 2a cos (2πx/λ) sin (2πt/T) how was the second step in terms of COS arrived?

y = y1+y2
= a [sin 2π (t/T – x/λ) + sin 2π (t/T + x/λ)]
= a [2sin (2πt/T) cos (2πx/λ)]
So, y = 2a cos (2πx/λ) sin (2πt/T)
how was the second step in terms of COS arrived?

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y = y 1 +y 2 = a [sin 2π (t/T – x/λ) + sin 2π (t/T + x/λ)] = a [2sin (2πt/T) cos (2πx/λ)] So, y = 2a cos (2πx/λ) sin (2πt/T) how was the second step in terms of COS arrived?

y = y1+y2= a [sin 2π (t/T – x/λ) + sin 2π (t/T + x/λ)]= a [2sin (2πt/T) cos (2πx/λ)]So, y = 2a cos (2πx/λ) sin (2πt/T) how was the second step in terms of COS arrived?

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y = y1+y2
= a [sin 2π (t/T – x/λ) + sin 2π (t/T + x/λ)]
= a [2sin (2πt/T) cos (2πx/λ)]
So, y = 2a cos (2πx/λ) sin (2πt/T)
how was the second step in terms of COS arrived?