If $A = 2(sin^6 θ + cos^6 θ)-3(sin^4θ+cos^4θ)$ then the value of 3α such that cos α = $\sqrt{\frac{3+A}{5+A}}$ is :
Answer & explanation
Correct answer: option 2
We know,
( sin6 θ + cos6 θ ) = 1 - 3 sin²θ. cos²θ
& ( sin4 θ + cos4 θ ) = 1 - 2 sin²θ. cos²θ
A = 2 ( sin6 θ + cos6 θ ) - 3 ( sin4 θ + cos4 θ )
= 2 × ( 1 - 3 sin²θ. cos²θ ) - 3 × ( 1 - 2 sin²θ. cos²θ )
= 2 - 6sin²θ. cos²θ -3 + 6 sin²θ. cos²θ
= - 1
Now,
cos α = \(\sqrt { \frac{3 + A}{5 + A }\ }\)
= \(\sqrt { \frac{3 - 1 }{5 - 1 }\ }\)
= \(\sqrt { \frac{1}{2 }\ }\)
{ we know, cos 45º = \(\sqrt { \frac{1}{2 }\ }\) }
So, α = 45º
⇒ 3 α = 3 × 45º
= 135º