Plz me with all the steps
prove that
sin^6(x)+cos^6 (x) = 1- 3/4 sin^2(2x)
i really need

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)

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