Why is sin ⁡ 8 t sin ⁡ 2 t = cos ⁡ 6 t − cos ⁡ 10 t 2 ?

If we write the functions sine and cosine for real numbers in exponential form they are easy to manipulate. We have that sin ⁡ ( x ) = e i x − e − i x 2 i and cos ⁡ ( x ) = e i x + e − i x 2 Then sin ⁡ ( 8 t ) sin ⁡ ( 2 t ) = e i 8 t − e − i 8 t 2 i ⋅ e i 2 t − e − i 2 t 2 i = − e i 10 t − e i 6 t − e − i 6 t + e − i 10 t 4 = = 1 2 ( e i 6 t + e − i 6 t 2 − e i 10 t + e − i 10 t 2 ) = 1 2 ( cos ⁡ ( 6 t ) − cos ⁡ ( 10 t ) )

"Why is sin ⁡ 8 t sin ⁡..." solved and explained

gnatopoditw

Answered question

2022-06-11

Why is

\sin 8 t \sin 2 t = \frac{\cos 6 t - \cos 10 t}{2}

Answer & Explanation

Braedon Rivas

Beginner2022-06-12Added 24 answers

If we write the functions sine and cosine for real numbers in exponential form they are easy to manipulate. We have that

\sin (x) = \frac{e^{i x} - e^{- i x}}{2 i} and \cos (x) = \frac{e^{i x} + e^{- i x}}{2}

Then

\sin (8 t) \sin (2 t) = \frac{e^{i 8 t} - e^{- i 8 t}}{2 i} \cdot \frac{e^{i 2 t} - e^{- i 2 t}}{2 i} = - \frac{e^{i 10 t} - e^{i 6 t} - e^{- i 6 t} + e^{- i 10 t}}{4} = = \frac{1}{2} (\frac{e^{i 6 t} + e^{- i 6 t}}{2} - \frac{e^{i 10 t} + e^{- i 10 t}}{2}) = \frac{1}{2} (\cos (6 t) - \cos (10 t))

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