Question

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

simplify it

Answer 1

Answer:

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

sin

2

(x)−cos

2

(x)

sin

4

(x)−cos

4

(x)

=1

Step-by-step explanation:

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

sin

2

(x)−cos

2

(x)

sin

4

(x)−cos

4

(x)

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

4

(x)−cos

4

(x):(sin

2

(x)+cos

2

(x))(sin(x)+cos(x))(sin(x)−cos(x))

\sin ^4(x)-\cos ^4(x)sin

4

(x)−cos

4

(x)

\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^cApplyexponentrule:a

bc

=(a

b

)

c

\sin ^4(x)=(\sin ^2(x))^2sin

4

(x)=(sin

2

(x))

2

=(\sin ^2(x))^2-\cos ^4(x)=(sin

2

(x))

2

−cos

4

(x)

\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^cApplyexponentrule:a

bc

=(a

b

)

c

\cos ^4(x)=(\cos ^2(x))^2cos

4

(x)=(cos

2

(x))

2

=(\sin ^2(x))^2-(\cos ^2(x))^2=(sin

2

(x))

2

−(cos

2

(x))

2

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

2

−y

2

=(x+y)(x−y)

(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))(sin

2

(x))

2

−(cos

2

(x))

2

=(sin

2

(x)+cos

2

(x))(sin

2

(x)−cos

2

(x))

=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))=(sin

2

(x)+cos

2

(x))(sin

2

(x)−cos

2

(x))

\mathrm{Factor}\:\sin ^2(x)-\cos ^2(x)Factorsin

2

(x)−cos

2

(x)

\sin ^2(x)-\cos ^2(x)sin

2

(x)−cos

2

(x)

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

2

−y

2

=(x+y)(x−y)

\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))sin

2

(x)−cos

2

(x)=(sin(x)+cos(x))(sin(x)−cos(x))

=(\sin (x)+\cos (x))(\sin (x)-\cos (x))=(sin(x)+cos(x))(sin(x)−cos(x))

=(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))=(sin

2

(x)+cos

2

(x))(sin(x)+cos(x))(sin(x)−cos(x))

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

sin

2

(x)−cos

2

(x)

(sin

2

(x)+cos

2

(x))(sin(x)+cos(x))(sin(x)−cos(x))

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

2

−y

2

=(x+y)(x−y)

\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))sin

2

(x)−cos

2

(x)=(sin(x)+cos(x))(sin(x)−cos(x))

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

(sin(x)+cos(x))(sin(x)−cos(x))

(sin

2

(x)+cos

2

(x))(sin(x)+cos(x))(sin(x)−cos(x))

\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)+\cos (x)Cancelthecommonfactor:sin(x)+cos(x)

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

sin(x)−cos(x)

(sin

2

(x)+cos

2

(x))(sin(x)−cos(x))

\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)-\cos (x)Cancelthecommonfactor:sin(x)−cos(x)

=\sin ^2(x)+\cos ^2(x)=sin

2

(x)+cos

2

(x)

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

2

(x)+sin

2

(x)=1

=1=1

Answer 2

$\color{red} {{{\Large {\bf{To\:\:Simplify\::\frac{\sin ^4\left(x\right)-\cos ^4\left(x\right)}{\sin ^2\left(x\right)-\cos ^2\left(x\right)}}}}}}$

$\color{green}{{{\large {\bf{Your\:\:Answer\::\frac{\sin ^4\left(x\right)-\cos ^4\left(x\right)}{\sin ^2\left(x\right)-\cos ^2\left(x\right)}=1}}}}}$

$\color{yellow} {\Huge {\sf{Solution:}}}$

$\color{blue} {\large {\bf{Factor\:\sin ^4\left(x\right)-\cos ^4\left(x\right)}}}$

$\tt \color{blue} {\mathrm{Rewrite\:}\sin ^4\left(x\right)-\cos ^4\left(x\right)\mathrm{\:as\:}\left(\sin ^2\left(x\right)\right)^2-\left(\cos ^2\left(x\right)\right)^2=\left(\sin ^2\left(x\right)\right)^2-\left(\cos ^2\left(x\right)\right)^2}$

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

$\color{fuchsia} {\normalsize \sin ^4\left(x\right)=\left(\sin ^2\left(x\right)\right)^2}$

$\color{fuchsia} {\normalsize =\left(\sin ^2\left(x\right)\right)^2-\cos ^4\left(x\right)}$

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

$\color{fuchsia} {\normalsize \cos ^4\left(x\right)=\left(\cos ^2\left(x\right)\right)^2}$

$\color{fuchsia} {\normalsize =\left(\sin ^2\left(x\right)\right)^2-\left(\cos ^2\left(x\right)\right)^2}$

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

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

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

$\color{blue} {\large {\bf{Factor\:\sin ^2\left(x\right)-\cos ^2\left(x\right)}}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\huge \boxed{\color{red} {\ \huge =1}}$