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

#### MarkFL

Staff member
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

Staff member
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.