Question

Prove that 2(sin⁶0+ cos⁶0)-3(sin⁴ 0+ cos⁴0) is equal to-1.​ plz answer quick plz​

Answer 1

ANSWER:

To Prove:

• 2(sin⁶θ + cos⁶θ) - 3(sin⁴θ + cos⁴θ) = -1

Proof:

We need to prove that,

$:\longrightarrow 2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)=-1$

Taking LHS,

$:\implies 2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta)$

$:\implies 2[(\sin^2\theta)^3+(\cos^2\theta)^3]-3[(\sin^2\theta)^2+(\cos^2\theta)^2]$

We know that,

$:\hookrightarrow a^3+b^3=(a+b)^3-3ab(a+b),\:and$

$:\hookrightarrow a^2+b^2=(a+b)^2-2ab$

So,

$:\implies 2[(\sin^2\theta)^3+(\cos^2\theta)^3]-3[(\sin^2\theta)^2+(\cos^2\theta)^2]$

Hence,

$:\implies 2[(\sin^2\theta+\cos^2\theta)^3-3(\sin^2\theta\cos^2\theta)(\sin^2\theta+\cos^2\theta)]-3[(\sin^2\theta+\cos^2\theta)^2-2(\sin^2\theta\cos^2\theta)]$

We know that,

$:\hookrightarrow \sin^2\theta+\cos^2\theta=1$

So,

$:\implies 2[(\sin^2\theta+\cos^2\theta)^3-3(\sin^2\theta\cos^2\theta)(\sin^2\theta+\cos^2\theta)]-3[(\sin^2\theta+\cos^2\theta)^2-2(\sin^2\theta\cos^2\theta)]$

Hence,

$:\implies 2[(1)^3-3(\sin^2\theta\cos^2\theta)(1)]-3[(1)^2-2(\sin^2\theta\cos^2\theta)]$

On simplifying,

$:\implies 2[1-3\sin^2\theta\cos^2\theta]-3[1-2\sin^2\theta\cos^2\theta]$

On opening brackets,

$:\implies 2-6\sin^2\theta\cos^2\theta-3+6\sin^2\theta\cos^2\theta$

On rearranging,

$:\implies 2-3+(-6\sin^2\theta\cos^2\theta+6\sin^2\theta\cos^2\theta)$

So, 6sin²θ cos²θ gets cancelled,

$:\implies -1+0$

$:\implies\bf-1=RHS$

HENCE PROVED!!!

Formulae Used:

• $a^3+b^3=(a+b)^3-3ab(a+b)$
• $a^2+b^2=(a+b)^2-2ab$
• $\sin^2\theta+\cos^2\theta=1$