if cosθ+cos∅=a and sinθ+sin∅=b than show that cos(θ+∅)= a²-b²/a²+b²

Question

Plz help me to prove this...

Answer 1

 

We need two identities as well as the sum to product and product to sum formulas.

 

The first identity is :

          (cosX)^2 - (sinX)^2 = 1 - (sinX)^2 - (sinX)^2 = 1 - 2 * sinX^2 = 1 - 2sinX^2

 

The second comes from the product to sum formula:

      (sinX)^2 = sinX*Sinx

                    = 1/2(  cos( X-X) - cos(X +X))

                     = 1/2( cos(0) - cos2X)

                     = 1/2( 1 - cos2X)

 

I would recommend you research product to sum and sum to product formulas online.

Peruse them thoroughly as they are used without much warning in step 4 of the proof.

 

--------------------------------------------------------------------------------------------------

 

The identity is proven right to left. I will denote the angles as T and X.

 

(cos T + cos X)^2 - ( sin T + sin X)^2

-------------------------------------------- =

 (cos T +cos X)^2 + ( sin T + sin X)^2

 

 

STEP 1: FOILing everything:

 

cosT^2 + 2cosTcosX + cos X^2 - sinT^2 - 2sinTsinX - sinX^2

--------------------------------------------------------------------------

cosT^2 + 2cosTcosX + cos X^2 + sinT^2 + 2sinTsinX + sinX^2

 

 

STEP 2:

Applying trig identity #1 from above to  the cosine^2 - sine^2 terms in the numerator;

Also applying sin^2 + cos^2 = 1 to the applicable terms in the denominator:

 

2 cosT cosX - 2 sinT sinX + 1 - 2 sin T^2 + 1 - 2 sin X^2

-------------------------------------------------------------------

 2 + 2cosT cos X + 2 sin T sin X

 

 

Step 3:  the ONEs (1s) in the numerator combine.

             A factor of 2 can be cancelled from everything

 

cos T cos X - sin T sin X +1 - sin T ^2 - sin X^2

-------------------------------------------------------

 1 + cosT cos X + sin T sin X

 

 

Step 4: Applying the product to sum formulas to all of the 

             cosT cosX and sinT sinX:

 

cos ( T + X ) + 1 - sin T ^2 - sin X^2

------------------------------------------- 

               1 + cos ( T - X)

 

 

Step 5: Applying the second identity above to the sine^2 terms in 

            the numerator

 

 cos (X + T) + 1 - 1/2 + cos 2T  - 1/2 + cos 2X

-----------------------------------------------------

              1 + cos ( T - X)

 

 

 

Step 6:  the One cancels the halves

 

cos (X + T) + cos 2T  + cos 2X
-----------------------------------------------------
1 + cos ( T - X)

 

 

Step 7: Applying the sum to product formula to

            the double angle cosine terms in the numerator

 

    cos ( T + X) + 1/2* 2 cos( T + X) cos (T - X)

   ---------------------------------------------------

          1 + cos (T - X)

 

Step 8: 1/2 * 2 cancels; cos (T + X) is factored out;

              denominator cancels with the numerator after factoring

 

      cos (T + X)  [1 + cos (T-x)]

       -------------------------------

                1 + cos(T-x)

 

    End of Proof

 

The tough part is getting the hang of the algebra behind the

product to sum and sum to product formulas,

This was a challenging problem. Thank you!