SOLUTION: Show that: {{{(1+sin(x)+icos(x))/(1+sin(x)-icos(x))}}} = sin(x) + icos(x)

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson  -> Lesson -> SOLUTION: Show that: {{{(1+sin(x)+icos(x))/(1+sin(x)-icos(x))}}} = sin(x) + icos(x)      Log On


   

Question 1152684: Show that: %281%2Bsin%28x%29%2Bicos%28x%29%29%2F%281%2Bsin%28x%29-icos%28x%29%29 = sin(x) + icos(x)
Answer by jim_thompson5910(35256) About Me  (Show Source):
You can put this solution on YOUR website!

The left hand side (LHS) is in the form %28A%2BB%29%2F%28A-B%29 where A+=+1%2Bsin%28x%29 and B+=+i%2Acos%28x%29

Throughout this entire problem, I will not alter the right hand side (RHS). When it comes to proving identities, the method is to pick one side and transform it into the other. Whatever side you havent picked to transform will remain the same the entire time.

Multiply top and bottom of the LHS by %28A%2BB%29%2F%28A%2BB%29 so that we transform the denominator into a real number

Original equation

Multiply top and bottom by %28A%2BB%29%2F%28A%2BB%29

Plug in values of A and B.

Numerator is a perfect square. Difference of squares rule in the denominator

Use the idea that i%5E2+=+-1

Pythagorean trig identity: sin^2+cos^2 = 1 (denominator LHS)

Combine like terms

Factor

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

Let's take a brief pause from that. We'll come back to it of course. The numerator of the LHS will get a bit ugly when we expand it out, so let's just focus on that for now.

We have something in the form %28A%2BB%29%5E2 with A+=+1%2Bsin%28x%29 and B+=+i%2Acos%28x%29

Use the rule that %28A%2BB%29%5E2+=+A%5E2+%2B+2AB+%2B+B%5E2

So,
%28A%2BB%29%5E2+=+A%5E2+%2B+2AB+%2B+B%5E2

Plug in the previously mentioned values of A and B

Use i^2 = -1

Expand (1+sin(x))^2

Use sin^2 = 1-cos^2 (variation of the pythagorean trig identity)

Combine like terms (pair of '1's, also pair of -cos^2 terms)

Factor out the GCF 2

Replace the term '1' with sin^2+cos^2, which helps with the canceling on the next step.

Combine like terms. Cos^2 terms cancel.

Pair up terms, then factor by grouping

Complete the factor by grouping process

Rearrange terms. We'll use this later. Call this equation (1).

There is possibly a more efficient method to expanding out this expression, but I can't think of it right now.

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

Back to the main event


Replace the numerator on the LHS using equation (1)

We have these pairs of terms

that cancel out

sin%28x%29%2Bi%2Acos%28x%29+=+sin%28x%29+%2B+i%2Acos%28x%29 The LHS has been transformed into the RHS. The identity is confirmed.