plz help me with all the stepsprove that sin^6(x)+cos^6 (x) = 1- 3/4 sin^2(2x)i really need help...​

Step-by-step explanation:Since the identity is true whether the angle x is measured in degrees, radians, gradians (indeed, anything else you care to concoct), I’ll omit the ‘degrees’ sign. Using the binomial theorem, (a+b)3=a3+3a2b+3ab2+b3 ⇒a3+b3=(a+b)3−3a2b−3ab2=(a+b)3−3(a+b)ab Substituting a=sin2(x) and b=cos2(x), we have: sin6(x)+cos6(x)=(sin2(x)+cos2(x))3−3(sin2(x)+cos2(x))sin2(x)cos2(x) Using the trigonometric identity cos2(x)+sin2(x)=1, your expression simplifies to: sin6(x)+cos6(x)=1−3sin2(x)cos2(x) From the double angle formula for the sine function, sin(2x)=2sin(x)cos(x)⇒sin(x)cos(x)=0.5sin(2x) Meaning the expression can be rewritten as: sin6(x)+cos6(x)=1−0.75sin2(2x)=1−34sin2(2x)