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

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To prove the trigonometric identity \( \frac{\sin^4(x)-\cos^4(x)}{\sin^3(x)+\cos^3(x)} = \frac{\sin(x)-\cos(x)}{1-\sin(x)\cos(x)} \), start with the left side and apply algebraic formulas. Factor the numerator using the difference of squares and the denominator using the sum of cubes. After simplifying, utilize the Pythagorean identity to rewrite the expression. Cancel common factors to arrive at the desired result, confirming the identity holds true. The steps demonstrate a clear application of trigonometric identities and algebraic manipulation.
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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,
 
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Hello Hello,

We are given to prove:

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

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:

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

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:

$$a^2-b^2=(a+b)(a-b)$$

ii) Sum of cubes:

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

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

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

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

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

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

Dividing out common factors, we see:

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

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

Shown as desired.
 
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