Question

If sin^6 θ + cos^6 θ + k cos^2 2θ = 1 and , where p and q are co-prime, then find p + q.

Answer 1

Answer:

Step-by-step explanation:We know that,

sin2 θ + cos2 θ = 1

Taking,

L.H.S = tan2 θ cos2 θ

= (tan θ × cos θ)2

= (sin θ)2

= sin2 θ

= 1 – cos2 θ

= R.H.S

– Hence Proved

Answer 2

GIVEN:-

$sin {}^{6} \alpha + cos {}^{6} \alpha + kcos {}^{2} 2 \alpha = 1$

REQUIRED TO FIND :- Value of k.

SOLUTION: -

We have,

sin^ 6 θ+cos^6 θ=1−ksin^2 2θ

☆ a^3+b^3=(a^2+b^2)^3−3a^2.b^2(a^2+b^2)☆

=(sin^2θ+cos^2θ)^3−3sin^2θ.cos^2θ(sin^2θ+cos^2θ)=1−ksin^22θ

13−3sin2θ.cos2θ=1−ksin22θ

k=sin22θ3sin2θ.cos2θ

=2sin2θ.cos2θ3sin2θ.cos2θ

Therefore,

》k=3/2

HOPE IT HELPS :)