Cos²x + cos² (x +1209) + cos² (x-120) = 2/3

1.	Cos²x + cos² (x +1209) + cos² (x-120) = 2/3
Answer» $\huge\mathcal{\fcolorbox{lime}{black}{\pink{<klux>ANSWER</klux>}}}$ $\color{red} {{{\Large {\bf{To\:\:Simplify\::\frac{\<klux>SIN</klux> ^4(<klux>X</klux>)-\<klux>COS</klux> ^4(x)}{\sin ^2(x)-\cos ^2(x)}}}}}}$ $\color{green}{{{\large {\bf{Your\:\:Answer\::\frac{\sin ^4(x)-\cos ^4(x)}{\sin ^2(x)-\cos ^2(x)}=1}}}}}$ $\color{yellow} {\Huge {\sf{Solution:}}}$ $\color{blue} {\large {\bf{Factor\:\sin ^4(x)-\cos ^4(x)}}}$ $\tt \color{blue} {\mathrm{Rewrite\:}\sin ^4(x)-\cos ^4(x)\mathrm{\:as\:}(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x))^2-(\cos ^2(x))^2}$ $\color{fuchsia} {\normalsize {\mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}}$ $\color{fuchsia} {\normalsize \sin ^4(x)=(\sin ^2(x))^2}$ $\color{fuchsia} {\normalsize =(\sin ^2(x))^2-\cos ^4(x)}$ = $\color{fuchsia} {\normalsize \mathrm{Apply\:exponent\:rule}:\quad \:a^{bc}=(a^b)^c}$ $\color{fuchsia} {\normalsize \cos ^4(x)=(\cos ^2(x))^2}$ $\color{fuchsia} {\normalsize =(\sin ^2(x))^2-(\cos ^2(x))^2}$ = $\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$ x^2-y^2=(x+y)(x-y) $(\sin ^2(x))^2-(\cos ^2(x))^2=(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))$ = $(\sin ^2(x)+\cos ^2(x))(\sin ^2(x)-\cos ^2(x))$ = $\color{blue} {\large {\bf{Factor\:\sin ^2(x)-\cos ^2(x)}}}$ $\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$ x^2-y^2=(x+y)(x-y) $\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)-\cos (x))$ (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)) $\large=(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))\ \textless \ br /\ \textgreater \ (x))(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ \large =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{\sin ^2(x)-\cos ^2(x)}$ = $\mathrm{Apply\:Difference\:of\:Two\:Squares\:Formula:\:}$ x^2-y^2=(x+y)(x-y) $\sin ^2(x)-\cos ^2(x)=(\sin (x)+\cos (x))(\sin (x)$ $(x)=(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}$ = $\mathrm{Cancel\:}\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}:\quad \sin ^2(x)+\cos ^2(x)Cancel \ \textless \ br /\ \textgreater \ (sin(x)+cos(x))(sin(x)−cos(x))$ $\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)+\cos(x)Cancelthecommonfactor:sin(x)+cos(x)$ = $\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)-\cos (x))}{\sin (x)-\cos (x)}$ = $\mathrm{Cancel\:the\:common\:factor:}\:\sin (x)-\cos$ $\mathrm{Use\:the\:following\:identity}:\quad \cos ^2(x)+\sin$ $\huge \boxed{\color{red} {\ \huge =1}}$ Answer: $\huge{Hope\;it\;helps\;u}$ Step-by-step EXPLANATION: $\huge{Hope\;it\;helps\;u}$

Answer»

$\huge\mathcal{\fcolorbox{lime}{black}{\pink{<klux>ANSWER</klux>}}}$

$\color{red} {{{\Large {\bf{To\:\:Simplify\::\frac{\<klux>SIN</klux> ^4(<klux>X</klux>)-\<klux>COS</klux> ^4(x)}{\sin ^2(x)-\cos ^2(x)}}}}}}$

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

(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))

$\large=(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))\ \textless \ br /\ \textgreater \ (x))(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ \large =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{\sin ^2(x)-\cos ^2(x)}$ =

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

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

$(x)=(sin(x)+cos(x))(sin(x)−cos(x))\ \textless \ br /\ \textgreater \ =\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}$ =

$\mathrm{Cancel\:}\frac{(\sin ^2(x)+\cos ^2(x))(\sin (x)+\cos (x))(\sin (x)-\cos (x))}{(\sin (x)+\cos (x))(\sin (x)-\cos (x))}:\quad \sin ^2(x)+\cos ^2(x)Cancel \ \textless \ br /\ \textgreater \ (sin(x)+cos(x))(sin(x)−cos(x))$