Trigonometric identity of finite termsProve that: 1 cos ⁡ x + cos ⁡ 3 x + 1 cos ⁡ x + cos ⁡ 5 x + ⋯ + 1 cos ⁡ x + cos ⁡ ( 2 n + 1 ) x = 1 2 csc ⁡ x [ tan ⁡ ( n + 1 ) x − tan ⁡ x ]I tried to prove this using the regular formulas. But failed. Please help me.

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Trigonometric identity of finite termsProve that:            1              cos        ⁡        x        +        cos        ⁡                  3          x                      +            1              cos        ⁡        x        +        cos        ⁡                  5          x                      +  ⋯  +            1              cos        ⁡        x        +        cos        ⁡                  (          2          n          +          1          )          x                        =      1    2    csc  ⁡  x    [  tan  ⁡      (    n    +    1    )    x    −  tan  ⁡      x    ]I tried to prove this using the regular formulas. But failed. Please help me.

Azul Lang · Accepted Answer

∑          k      =      1        n        1          cos      ⁡              x            +      cos      ⁡      (      2      k      +      1      )      x        =      ∑          k      =      1        n        1          2      cos      ⁡              k        x            cos      ⁡      (      k      +      1      )      x        =  =      1          2      sin      ⁡              x                  ∑          k      =      1        n        (    tan    ⁡    (    k    +    1    )    x    −    tan    ⁡    k    x    )    =      1          2      sin      ⁡              x                  (    tan    ⁡    (    n    +    1    )    x    −    tan    ⁡    x    )  and we are done!I used the following reasoning.  tan  ⁡  α  −  tan  ⁡  β  =            sin      ⁡      α              cos      ⁡      α        −            sin      ⁡      β              cos      ⁡      β        =            sin      ⁡      α      cos      ⁡      β      −      cos      ⁡      α      sin      ⁡      β              cos      ⁡      α      cos      ⁡      β        =            sin      ⁡      (      α      −      β      )              cos      ⁡      α      cos      ⁡      β        .

genestesya · Accepted Answer

In the case n=0 there is no summand on the left-hand side and the right-hand side is 0, so the base case of the induction holds.Suppose it holds for n−1; then the left-hand side with n can be written, by the induction hypothesis,      1          2      sin      ⁡      x                  (        tan  ⁡  n  x  −  tan  ⁡  x            )        +      1          cos      ⁡      x      +      cos      ⁡      (      2      n      +      1      )      x      and you want to prove this equals      1          2      sin      ⁡      x                  (        tan  ⁡  (  n  +  1  )  x  −  tan  ⁡  x            )      which is equivalent to            tan      ⁡      n      x              2      sin      ⁡      x        +      1          cos      ⁡      x      +      cos      ⁡      (      2      n      +      1      )      x        =            tan      ⁡      (      n      +      1      )      x              2      sin      ⁡      x      By the sum-to-product formulas, this becomes            sin      ⁡      n      x              2      sin      ⁡      x      cos      ⁡      n      x        +      1          2      cos      ⁡      n      x      cos      ⁡      (      n      +      1      )      x        =            sin      ⁡      (      n      +      1      )      x              2      sin      ⁡      x      cos      ⁡      (      n      +      1      )      x      Reduce to the same denominator and conclude the equality holds.The relation  sin  ⁡  n  x  cos  ⁡  (  n  +  1  )  x  +  sin  ⁡  x  =  sin  ⁡  (  n  +  1  )  x  cos  ⁡  n  xis true, because it is equivalent to  sin  ⁡  x  =  sin  ⁡  (  n  +  1  )  x  cos  ⁡  n  x  −  cos  ⁡  (  n  +  1  )  x  sin  ⁡  n  x

Trigonometric identity of finite terms Prove that: 1/(cos x+cos 3x) + 1/(cos x+ cos (5x))+...+1/(cos x+ cos ((2n+1)x))=1/2 csc x [ tan((n+1)x)-\tan(x)] I tried to prove this using the regular formulas. But failed. Please help me.

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