Show that 1 + sin ⁡ A cos ⁡ A + cos ⁡ B 1 − sin ⁡ B = 2 sin ⁡ A − 2 sin ⁡ B sin ⁡ ( A − B ) + cos ⁡ A − cos ⁡ B

Question

Show that             1      +      sin      ⁡      A              cos      ⁡      A        +            cos      ⁡      B              1      −      sin      ⁡      B        =            2      sin      ⁡      A      −      2      sin      ⁡      B              sin      ⁡      (      A      −      B      )      +      cos      ⁡      A      −      cos      ⁡      B

soymmernenx · Accepted Answer

The LHS can be written as                     1        +        sin        ⁡        A                    cos        ⁡        A              +                    1        +        sin        ⁡        B                    cos        ⁡        B              =                    cos        ⁡        A        +        cos        ⁡        B        +        sin        ⁡        A        cos        ⁡        B        +        cos        ⁡        A        sin        ⁡        B                    cos        ⁡        A        cos        ⁡        B            Multiply Neumaretor and Denominator by   sin  ⁡  (  A  −  B  )  +  cos  ⁡  A  −  cos  ⁡  B  =  (  cos  ⁡  A  +  sin  ⁡  A  cos  ⁡  B  )  −  (  cos  ⁡  B  +  cos  ⁡  A  sin  ⁡  B  )  =                    (        cos        ⁡        A        +        sin        ⁡        A        cos        ⁡        B                  )          2                −        (        cos        ⁡        B        +        cos        ⁡        A        sin        ⁡        B                  )          2                            cos        ⁡        A        cos        ⁡        B        [        .        .        .        ]              =                              cos          2                ⁡        A        +                  sin          2                ⁡        A                  cos          2                ⁡        B        +        2        sin        ⁡        A        cos        ⁡        A        cos        ⁡        B        −                  cos          2                ⁡        B        −                  cos          2                ⁡        A                  sin          2                ⁡        B        −        2        cos        ⁡        A        cos        ⁡        B        sin        ⁡        B                    cos        ⁡        A        cos        ⁡        B        [        .        .        .        ]            Note that       sin    2    ⁡  A      cos    2    ⁡  B  −      cos    2    ⁡  A      sin    2    ⁡  B  =  (  1  −      cos    2    ⁡  A  )      cos    2    ⁡  B  −      cos    2    ⁡  A  (  1  −      cos    2    ⁡  B  )  =      cos    2    ⁡  B  −      cos    2    ⁡  ASo LHS becomes                    2        sin        ⁡        A        cos        ⁡        A        cos        ⁡        B        −        2        cos        ⁡        A        cos        ⁡        B        sin        ⁡        B                    cos        ⁡        A        cos        ⁡        B        [        .        .        .        ]              =                    2        sin        ⁡        A        −        2        sin        ⁡        B                    sin        ⁡        (        A        −        B        )        +        cos        ⁡        A        −        cos        ⁡        B              =  R  H  S

Show that 1 + sin ⁡ A </mrow>

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