I've been trying to figure out why this equation is satisfied: 1 − ( cos ⁡ ( x ) ) 3 x 2 = 2 ⋅ ( sin ⁡ ( x 2 ) ) 2 x 2 ⋅ ( 1 + cos ⁡ ( x ) + ( cos ⁡ ( x ) ) 2 )but I can't find the proper trigonometric changes in order to change from one to another. I know that the sine comes from the double angle formula, but I obtain slightly different results in other parts and it's never the same as the formula above.

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I&#039;ve been trying to figure out why this equation is satisfied:            1      −      (      cos      ⁡      (      x      )              )        3                    x      2        =            2      ⋅      (      sin      ⁡      (              x        2            )              )        2                    x      2        ⋅  (  1  +  cos  ⁡  (  x  )  +  (  cos  ⁡  (  x  )      )    2    )but I can&#039;t find the proper trigonometric changes in order to change from one to another. I know that the sine comes from the double angle formula, but I obtain slightly different results in other parts and it&#039;s never the same as the formula above.

floygdarvn · Accepted Answer

One may observe that, in general,      a    3    −      b    3    =  (  a  −  b  )  (      a    2    +  a  b  +      b    2    )giving  1  −  (  cos  ⁡  x      )    3    =  (  1  −  cos  ⁡  x  )  (  1  +  cos  ⁡  x  +  (  cos  ⁡  x      )    2    )then, by using   2      sin    2    ⁡  (  x      /    2  )  =  1  −  cos  ⁡  x as noticed by the &quot;double-angle&quot; formula   2      sin    2    ⁡  (  x      /    2  )  =  1  −  cos  ⁡  x

I've been trying to figure out why this equation is satisfied: 1 −

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