Use an expression for sin ⁡ ( 5 θ ) sin ⁡ ( θ ) to find the roots of the equation x 4 − 3 x 2 + 1 = 0 in trigonometric form

Question

Use an expression for             sin      ⁡      (      5      θ      )              sin      ⁡      (      θ      )       to find the roots of the equation       x    4    −  3      x    2    +  1  =  0 in trigonometric form

rynosluv101swv2s · Accepted Answer

HINT:Using Prosthaphaeresis Formula,  sin  ⁡  5  x  −  sin  ⁡  x  =  2  sin  ⁡  2  x  cos  ⁡  3  x  =  4  sin  ⁡  x  cos  ⁡  x  cos  ⁡  3  xIf   sin  ⁡  x  ≠  0  ,,                    sin        ⁡        5        x                    sin        ⁡        x              −  1  =  4  cos  ⁡  x  cos  ⁡  3  x  =  4  cos  ⁡  x  (  4      cos    3    ⁡  x  −  3  cos  ⁡  x  )  =  (  4      cos    2    ⁡  x      )    2    −  3  (  4      cos    2    ⁡  x  )OR replace       sin    2    ⁡  x with   1  −      cos    2    ⁡  x in your  5      cos    4    ⁡  x  −  10      cos    2    ⁡  x      sin    2    ⁡  x  +      sin    4    ⁡  xNow if   sin  ⁡  5  x  =  0  ,  5  x  =  n  π where n is any integer  x  =                    n        π            5       where   n  ≡  0  ,  ±  1  ,  ±  2    (  mod    5  )So, the roots of                     sin        ⁡        5        x                    sin        ⁡        x              =  0    ⟹    x  =                    n        π            5       where   n  ≡  ±  1  ,  ±  2    (  mod    5  )But                    sin        ⁡        5        x                    sin        ⁡        x              =  (  4      cos    2    ⁡  x      )    2    −  3  (  4      cos    2    ⁡  x  )  +  1

Use an expression for sin ⁡ ( 5 &#x03

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