Question

Question no. 17
Plz help me fast because i have my paper after one day

Answer 1

Take sin²θ+ cos²θ = 1

2(sin⁶θ+cos⁶θ)-3(sin⁴θ+cos⁴θ)+1

= 2{(sin²θ)³+(cos²θ)³}-3{(sin²θ)²+(cos²θ)²}+1

=2{(sin²θ+cos²θ)³-3sin²θcos²θ(sin²θ+cos²θ)}-3{(sin²θ+cos²θ)²-2sin²θcos²θ}+1

=2(1-3sin²θcos²θ)-3(1-2sin²θcos²θ)+1

=2-6sin²θcos²θ-3+6sin²θcos²θ+1

=-1+1

=0

Answer 2

Answer:

To Show: $2(sin^6\,\theta+cos^6\,\theta)-3(sin^4\,\theta+cos^4\,\theta)+1=0$

We use the following identities,

$sin^2\,\theta+cos^2\,\theta=1$

$a^3+b^3=(a+b)^3-3ab(a+b)$

$a^2+b^2=(a+b)^2-2ab$

Consider,

$2(sin^6\,\theta+cos^6\,\theta)-3(sin^4\,\theta+cos^4\,\theta)+1$

$\implies2((sin^2\,\theta)^3+(cos^2\,\theta)^3)-3((sin^2\,\theta)^2+(cos^2\,\theta)^2)+1$

$\implies2((sin^2\,\theta+cos^2\,\theta)^3-3sin^2\,\theta\:cos^2\,\theta(sin^2\,\theta+cos^2\,\theta))-3((sin^2\,\theta+cos^2\,\theta)^2-2\:sin^2\,\theta\:cos^2\,\theta)+1$

$\implies2((1)^3-3sin^2\,\theta\:cos^2\,\theta(1))-3((1)^2-2\:sin^2\,\theta\:cos^2\,\theta)+1$

$\implies2-6sin^2\,\theta\:cos^2\,\theta-3+6\:sin^2\,\theta\:cos^2\,\theta+1$

$\implies2-3+1$

$\implies0$

Hence Proved