Hello's question at Yahoo Answers regarding proving a trigonometric identity

[MarkFL]

Here is the question:

Solving Trig Identity?


sin^4x - cos^4x / sin^3x + cos^3x = sinx - cosx / 1 - sinxcosx

please explain the steps to me?

I have posted a link there to this topic so the OP can see my work,

 

[MarkFL]

Hello Hello,

We are given to prove:

\(\displaystyle \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\sin(x)-\cos(x)}{1-\sin(x)\cos(x)}\)

Traditionally, we begin with the left side and try to apply well-known algebraic formulas and trigonometric identities to obtain the right side. So let's look at the left side:

\(\displaystyle \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}\)

Now, it we factor the numerator as the difference of squares, and the denominator as the sum of cubes, we may write:

i) Difference of squares:

\(\displaystyle a^2-b^2=(a+b)(a-b)\)

ii) Sum of cubes:

\(\displaystyle a^3+b^3=(a+b)\left(a^2-ab+b^2 \right)\)

\(\displaystyle \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\left(\sin^2(x)+\cos^2(x) \right)\left(\sin^2(x)-\cos^2(x) \right)}{\left(\sin(x)+\cos(x) \right)\left(\sin^2(x)-\sin(x)\cos(x)+\cos^2(x) \right)}\)

Applying the Pythagorean identity \(\displaystyle \sin^2(\theta)+\cos^2(\theta)=1\) we have:

\(\displaystyle \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\sin^2(x)-\cos^2(x)}{\left(\sin(x)+\cos(x) \right)\left(1-\sin(x)\cos(x) \right)}\)

Factoring the numerator as the difference of squares, we find:

\(\displaystyle \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\left(\sin(x)+\cos(x) \right)\left(\sin(x)-\cos(x) \right)}{\left(\sin(x)+\cos(x) \right)\left(1-\sin(x)\cos(x) \right)}\)

Dividing out common factors, we see:

\(\displaystyle \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}= \frac{\cancel{\left(\sin(x)+\cos(x) \right)} \left(\sin(x)-\cos(x) \right)}{\cancel{\left(\sin(x)+ \cos(x) \right)}\left(1-\sin(x)\cos(x) \right)}\)

\(\displaystyle \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)}=\frac{\sin(x)-\cos(x)}{1-\sin(x)\cos(x)}\)

Shown as desired.