Question

Prove that:

$3(sinx-cosx{)}^4+6(sinx+cosx{)}^2+4(si{n}^{6}x+co{s}^{6}x)=13$

Answer 1

LHS = $3(sinx-cosx{)}^4+6(sinx+cosx{)}^2+4(si{n}^{6}x+co{s}^{6}x)$

$=3[si{n}^{4}x-4si{n}^{3}xcosx+6si{n}^{2}x\times co{s}^{2}x-4sinxco{s}^{3}x+co{s}^{4}x]+6[si{n}^{2}x+2sinxcosx+co{s}^{2}x]+4(si{n}^{6}x+co{s}^{6}x)$

$[\because (a-b{)}^4=a^4-4a^{3}b+6a^2{b}^2-4a{b}^3+b^2\text{by binomial expainsion}]$

$=3[si{n}^{4}x+co{s}^{4}x-4sinxcosx(si{n}^{2}x+co{s}^{2}x)+6si{n}^{2}xco{s}^{2}x]+6[1+2sinxcosx]+4\left[\right(co{s}^{2}x+si{n}^{2}x\left)\right(co{s}^{4}x-co{s}^{2}xsi{n}^{2}x+si{n}^{4}x\left)\right][\because a^3+b^3=(a+b\left)\right(a^2-ab+b^2\left)\right]=7[si{n}^{4}x+co{s}^{4}x]+18si{n}^{2}xco{s}^{2}x-4si{n}^{2}xco{s}^{2}x+6=7[si{n}^{4}x+co{s}^{4}x+2si{n}^{2}xco{s}^{2}x]+6=7[si{n}^{2}x+co{s}^{2}x{]}^2+6$
= 7 + 6 = 13
= RHS