Trigonometric Identities

In mathematics, trigonometric identities are equalities that involve trigonometric functions and are true for every single value of the occurring variables (see Identity (mathematics)). Geometrically, these are identities involving certain functions of one or more angles. These are distinct from triangle identities, which are identities involving both angles and side lengths of a triangle. Only the former are covered in this article.

                                                                                                                                            — Source Wikipedia.

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Proving Sum and Difference Identities of Sine:

To prove: (i) sine (A + B) = sine A cosine B + cosine A sine B, and

                (ii) sine (A − B) = sine A cosine B − cosine A sine B.

• Proving, sine (A + B) = sine A cosine B + cosine A sine B.
  
  Proof: We know that, sine (θ) = cosine (`pi/2` − θ)
            If θ = A + B, then we has,
                      sine (A + B) = cosine [`pi/2` − (A + B)]
  
                                         = cosine [`pi/2` − A − B)]
            Next, we rearrange the angles inside the cosine term, we get,
                      cosine [`pi/2` − A − B)] = cosine [(`pi/2` − A) − B]
  
            By using the cosine of the difference identities,
                      cosine (A − B) = cosine A cosine B + sine A sine B], we have,
                      cosine [(`pi/2` − A) − B] = cosine (`pi/2` − A) cosine (B) + sine (`pi/2` − A) sine (B)
  
               = sine A cosine B + cosine A sine B
  
            Since we know that, cosine (`pi/2` − A) = sine A, and
  
                                           sine (`pi/2` − A) = cosine A.      Therefore, sine (A + B) = sine A cosine B + cosine A sine B.
  Hence proved that, sine (A + B) = sine A cosine B + cosine A sine B.

• Proving, sine (A – B) = sine A cosine B – cosine A sine B.
  
  Proof: We know that, sine (A + B) = sine A cosine B + cosine A sine B.
  
            Replacing B with (−B), these identities becomes,
  
            sine (A + (− B)) = sine A cosine (− B) + cosine A sine (− B)
  
            Since we know that, cosine (−A) = cosine A, and
  
                                           sine (−A) = − sine A.
            sine (A − B) = sine A cosine B − cosine A sine B.Hence proved that, sine (A – B) = sine A cosine B – cosine A sine B.
  
  
  

Proving Identities of Sine Double Angle:

To prove: sine 2A = sine A cosine A.

Proof: We know that, sine (A + B) = sine A cosine B + cosine A sine B,

           we will use these identities to obtain the sine of a double angle.

           Let us consider, L.H.S. = sine (A + B),

           Now replace, B with A, we get,

                      sine (A + B) = sine (A + A)

                                         = sine 2 A

                      R.H.S. = sine A cosine B + cosine A sine B

           Now replace, B with A, we get,

                      Sine A cosine A + cosine A sine A = 2sine A cosine A.

           Since, L.H.S. = R.H.S.

     Hence proved that, sine 2A = 2sine A cosine A.

Proving Identities of Sine Half Angle:

To Prove: sine A = `sqrt((1-cos2A) / 2)` .

Proof: We know that, cos2A = cos² A − sin² A.

                                => cos2A = (1 − sin² A) − sin² A                           [where, cos² θ = 1 − sin² θ]

                                => cos2A = 1 − 2sin² A

                                => 2sin² A = 1 − cos2A

                                => sin²A = ` (1 - cos2A) / 2` .

                     Taking square root we get,

                                => sin A = `sqrt((1-cos2A) / 2)` .

            Hence proved that, sin A = `sqrt ((1-cos2A) / 2)` .

Summary of Proving Identities of Sine:

• Sum and Difference Identities:
  Sine (A + B) = sine A cosine B + cosine A sine B,
  
  Sine (A − B) = sine A cosine B − cosine A sine B.
• Double Angle Identities:Sine 2A = sine A cosine A.

• Half Angle Identities:
  
  Sine A = `sqrt((1-cos2A) / 2)` .