Using trigonometry, one can easily establish that cos ⁡ ( ω 1 t ) + cos ⁡ ( ω 2 t ) = 2 cos ⁡ [ ( ω 1 − ω 2 ) t / 2 ] cos ⁡ [ ( ω 1 + ω 2 ) t / 2 ] But can we exploit exp ⁡ ( i θ ) = cos ⁡ θ + i sin ⁡ θto arrive at the same answer using complex exponential form, i.e. starting from: exp ⁡ ( i ω 1 t ) + exp ⁡ ( i ω 2 t )and WITHOUT USING ANY TRIGONOMETRIC FORMULA?

Question

Using trigonometry, one can easily establish that  cos  ⁡  (      ω    1    t  )  +  cos  ⁡  (      ω    2    t  )  =  2  cos  ⁡      [    (          ω      1        −          ω      2        )    t          /        2    ]    cos  ⁡      [    (          ω      1        +          ω      2        )    t          /        2    ]  But can we exploit  exp  ⁡  (  i  θ  )  =  cos  ⁡  θ  +  i  sin  ⁡  θto arrive at the same answer using complex exponential form, i.e. starting from:  exp  ⁡  (  i      ω    1    t  )  +  exp  ⁡  (  i      ω    2    t  )and WITHOUT USING ANY TRIGONOMETRIC FORMULA?

atoandro8f04v · Accepted Answer

HINT:      e          2      i      A        +      e          2      i      B        =      e          i      (      A      +      B      )      )            (          e              i        (        A        −        B        )        )              +          e              −        i        (        A        −        B        )        )              )  Now use Euler identity       e          i      x        =  cos  ⁡  x  +  i  sin  ⁡  x and subsequently,      e          i      (      A      −      B      )      )        +      e          −      i      (      A      −      B      )      )        =  2  cos  ⁡  (  A  −  B  )

Using trigonometry, one can easily establish that cos ⁡ ( &#x0

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