Question

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

Answer 1

Answer:

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

Step-by-step explanation:

$\dfrac{\sin ^4\left(x\right)-\cos ^4\left(x\right)}{\sin ^2\left(x\right)-\cos ^2\left(x\right)}$

$\mathrm{Factor}\:\sin ^4\left(x\right)-\cos ^4\left(x\right):\quad \left(\sin ^2\left(x\right)+\cos ^2\left(x\right)\right)\left(\sin \left(x\right)+\cos \left(x\right)\right)\left(\sin \left(x\right)-\cos \left(x\right)\right)$

$\sin ^4\left(x\right)-\cos ^4\left(x\right)$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=\left(a^b\right)^c$

$\sin ^4\left(x\right)=\left(\sin ^2\left(x\right)\right)^2$

$=\left(\sin ^2\left(x\right)\right)^2-\cos ^4\left(x\right)$

$\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=\left(a^b\right)^c$

$\cos ^4\left(x\right)=\left(\cos ^2\left(x\right)\right)^2$

$=\left(\sin ^2\left(x\right)\right)^2-\left(\cos ^2\left(x\right)\right)^2$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\left(\sin ^2\left(x\right)\right)^2-\left(\cos ^2\left(x\right)\right)^2=\left(\sin ^2\left(x\right)+\cos ^2\left(x\right)\right)\left(\sin ^2\left(x\right)-\cos ^2\left(x\right)\right)$

$=\left(\sin ^2\left(x\right)+\cos ^2\left(x\right)\right)\left(\sin ^2\left(x\right)-\cos ^2\left(x\right)\right)$

$\mathrm{Factor}\:\sin ^2\left(x\right)-\cos ^2\left(x\right)$

$\sin ^2\left(x\right)-\cos ^2\left(x\right)$

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\sin ^2\left(x\right)-\cos ^2\left(x\right)=\left(\sin \left(x\right)+\cos \left(x\right)\right)\left(\sin \left(x\right)-\cos \left(x\right)\right)$

$=\left(\sin \left(x\right)+\cos \left(x\right)\right)\left(\sin \left(x\right)-\cos \left(x\right)\right)$

$=\left(\sin ^2\left(x\right)+\cos ^2\left(x\right)\right)\left(\sin \left(x\right)+\cos \left(x\right)\right)\left(\sin \left(x\right)-\cos \left(x\right)\right)$

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

$\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}x^2-y^2=\left(x+y\right)\left(x-y\right)$

$\sin ^2\left(x\right)-\cos ^2\left(x\right)=\left(\sin \left(x\right)+\cos \left(x\right)\right)\left(\sin \left(x\right)-\cos \left(x\right)\right)$

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

$\mathrm{Cancel\:the\:common\:factor:}\:\sin \left(x\right)+\cos \left(x\right)$

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

$\mathrm{Cancel\:the\:common\:factor:}\:\sin \left(x\right)-\cos \left(x\right)$

$=\sin ^2\left(x\right)+\cos ^2\left(x\right)$

$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)+\sin ^2\left(x\right)=1$

$=1$