Why squaring the trigonometric equation changes the solution?I have a trigonometric equation that is defined as: sin ⁡ ( α ) − cos ⁡ ( α ) = 1 2 Solving this equation by mathematica will yield α = 65.70 and α = − 155.705But As I solve it analytically, I will obtain different results:First I exponentiate both sides to the power of two: ( sin ⁡ ( α ) − cos ⁡ ( α ) ) 2 = 1 2 2 Now I expand the expressions: sin ⁡ ( α ) 2 − 2 sin ⁡ ( α ) cos ⁡ ( α ) + cos ⁡ ( α ) 2 = 1 4 As sin ⁡ ( α ) 2 + cos ⁡ ( α ) 2 = 1, I will have: 1 − 2 sin ⁡ ( α ) cos ⁡ ( α ) = 1 4 Again if I put 2 sin ⁡ ( α ) cos ⁡ ( α ) = sin ⁡ ( 2 α ) I will be left with: sin ⁡ ( 2 α ) = 3 4 Which will readily give α = 24.3So, why I am having different results? I don't understand.

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Why squaring the trigonometric equation changes the solution?I have a trigonometric equation that is defined as:  sin  ⁡  (  α  )  −  cos  ⁡  (  α  )  =      1    2  Solving this equation by mathematica will yield   α  =  65.70 and   α  =  −  155.705But As I solve it analytically, I will obtain different results:First I exponentiate both sides to the power of two:  (  sin  ⁡  (  α  )  −  cos  ⁡  (  α  )      )    2    =            1      2        2  Now I expand the expressions:  sin  ⁡  (  α      )    2    −  2  sin  ⁡  (  α  )  cos  ⁡  (  α  )  +  cos  ⁡  (  α      )    2    =      1    4  As   sin  ⁡  (  α      )    2    +  cos  ⁡  (  α      )    2    =  1, I will have:  1  −  2  sin  ⁡  (  α  )  cos  ⁡  (  α  )  =      1    4  Again if I put   2  sin  ⁡  (  α  )  cos  ⁡  (  α  )  =  sin  ⁡  (  2  α  ) I will be left with:  sin  ⁡  (  2  α  )  =      3    4  Which will readily give   α  =  24.3So, why I am having different results? I don&#039;t understand.

Milton Gilmore · Accepted Answer

If you square, you also get the solutions of  sin  ⁡  α  −  cos  ⁡  α  =  −      1    2  In general, if you have an equation of the form   f  (  x  )  =  g  (  x  ) and square both sides, you get, after rearranging,                    (*)                    f        (        x                  )          2                −        g        (        x                  )          2                =        0            that can be rewritten  (  f  (  x  )  −  g  (  x  )  )  (  f  (  x  )  +  g  (  x  )  )  =  0so the solutions of (*) are the solutions of   f  (  x  )  −  g  (  x  )  =  0 (the original equation) together with the solutions of   f  (  x  )  +  g  (  x  )  =  0A safer way to solve your equation is to set   t  =  tan  ⁡  (  α      /    2  ), so the equation becomes            2      t              1      +              t        2              −            1      −              t        2                    1      +              t        2              =      1    2  that reduces to      t    2    +  4  t  −  3  =  0with solutions  −  2  +      7      or    −  2  −      7  The first solutions corresponds to  α  =  2  arctan  ⁡  (      7    −  2  )  ≈      65.7    ∘  and the second one to  α  =  −  2  arctan  ⁡  (      7    +  2  )  ≈  −      155.7    ∘  or   ≈      204.3    ∘   if you want a value between   0 and   360.

MISA6zh · Accepted Answer

Squaring is often a useful (sometimes unavoidable) way to solve an equation, but it always introduces new solutions which may not fit the original problem. When you square both sides of an equation, you should always check your solutions to make sure they are applicable.If you can avoid squaring, you might try doing so so as not to be confused with extra solutions. For this particular problem, you can solve it without squaring as follows:            1      2           =     sin  ⁡  α  −  cos  ⁡  α     =         2      (  cos  ⁡      45    ∘    sin  ⁡  α  −  sin  ⁡      45    ∘    cos  ⁡  α  )     =         2    sin  ⁡  (  α  −      45    ∘    )which gives   sin  ⁡  (  α  −      45    ∘    )  =      1          2              2             which you can solve to give you the values obtained from Mathematica.

Why squaring the trigonometric equation changes the solution? I have a trigonometric equation that is defined as: sin(alpha)−cos(alpha)=1/2

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