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