Question

sin⁴q-cos⁴q=sin²q-cos²q

Answer 1

Answer:

$\mathrm{Prove\:}\sin ^4\left(q\right)-\cos ^4\left(q\right)=\sin ^2\left(q\right)-\cos ^2\left(q\right):\quad \mathrm{True}$

Step-by-step explanation:

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

$\gray{\mathrm{Manipulating\:left\:side}}$

$\gray{\sin ^4\left(q\right)-\cos ^4\left(q\right)}$

$\black{\mathrm{Factor}\:-\cos ^4\left(q\right)+\sin ^4\left(q\right):}$

$-\cos ^4\left(q\right)+\sin ^4\left(q\right)$

$\gray{\mathrm{Rewrite\:}\sin ^4\left(q\right)-\cos ^4\left(q\right)\mathrm{\:as\:}\left(\sin ^2\left(q\right)\right)^2-\left(\cos ^2\left(q\right)\right)^2}$

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

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

$\gray{\left(\sin ^2\left(q\right)\right)^2-\left(\cos ^2\left(q\right)\right)^2=\left(\sin ^2\left(q\right)+\cos ^2\left(q\right)\right)\left(\sin ^2\left(q\right)-\cos ^2\left(q\right)\right)}$

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

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

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

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

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

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

$\black{\mathrm{Expand}\:\left(-\cos \left(q\right)+\sin \left(q\right)\right)\left(\cos \left(q\right)+\sin \left(q\right)\right):}$

$\left(-\cos \left(q\right)+\sin \left(q\right)\right)\left(\cos \left(q\right)+\sin \left(q\right)\right)$

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

$\gray{a=\sin \left(q\right),\:b=\cos \left(q\right)}$

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

$\gray{\mathrm{We\:showed\:that\:the\:two\:sides\:could\:take\:the\:same\:form}}$

$\Rightarrow \mathrm{True}$

Answer 2

$\underline{\textbf{\large{To prove: }}}$

$\sf \underline{sin^4q-cos^4q=sin^2q-cos^2q}$

$\underline{\textbf{\large{Proof:}}}$

LHS = $\sf sin^4q - cos^4q$

$=\sf (\sin{}^{2} q) ^{2} - ({ \cos }^{2}q ){}^{2}$

As we know the identity,

$[\sf( a^2 - b^2 ) = (a + b) (a - b)]$

$=\sf ( { \sin}^{2}q + { \cos }^{2}q)( { \sin}^{2}q - { \cos }^{2}q)$

we know ,$[\sf\ sin^2(theta)+ cos^2(theta) = 1]$

therefore,

$=\sf (1)( { \sin}^{2}q - { \cos }^{2}q)$

$=\sf { \sin }^{2}q - { \cos }^{2}q$

= RHS

hence proved.