If sin θ = cos2 ⁡ θ , then the value of cos12 θ + 3 cos10 ⁡ θ + 3 cos8 ⁡ θ + cos6 ⁡ θ - 1 is equal to

Question

If sin θ = cos2 ⁡ θ , then the value of cos12 θ + 3 cos10 ⁡ θ + 3 cos8 ⁡ θ + cos6 ⁡ θ - 1 is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

sin θ= cos 2 θ ⇒ sin θ+ sin 2 θ=1 Now, cos 6 θ( cos 6 θ+3 cos 4 θ+3 cos 2 θ+1)-1 = cos 6 θ( cos 2 θ+1)3-1 =( cos 2 θ)3( sin θ+1)3-1 = sin 3 θ( sin θ+1)3-1 =( sin 2 θ+ sin θ)3-1=1-1=0

Q. If $s in θ = co s^{2} θ$ , then the value of $co s^{12} θ + 3 co s^{10} θ + 3 co s^{8} θ + co s^{6} θ - 1$ is equal to

$sin θ = cos^{2} θ \Rightarrow sin θ + sin^{2} θ = 1$
Now, $cos^{6} θ (cos^{6} θ + 3 cos^{4} θ + 3 cos^{2} θ + 1) - 1$
$= cos^{6} θ (cos^{2} θ + 1)^{3} - 1$
$= (cos^{2} θ)^{3} (sin θ + 1)^{3} - 1$
$= sin^{3} θ (sin θ + 1)^{3} - 1$
$= (sin^{2} θ + sin θ)^{3} - 1 = 1 - 1 = 0$

Q. If sinθ=cos2⁡θ , then the value of cos12θ+3cos10⁡θ+3cos8⁡θ+cos6⁡θ−1 is equal to

sinθ=cos2θ⇒sinθ+sin2θ=1 Now, cos6θ(cos6θ+3cos4θ+3cos2θ+1)−1 =cos6θ(cos2θ+1)3−1 =(cos2θ)3(sinθ+1)3−1 =sin3θ(sinθ+1)3−1 =(sin2θ+sinθ)3−1=1−1=0

Q. If $s in θ = co s^{2} θ$ , then the value of $co s^{12} θ + 3 co s^{10} θ + 3 co s^{8} θ + co s^{6} θ - 1$ is equal to

$sin θ = cos^{2} θ \Rightarrow sin θ + sin^{2} θ = 1$
Now, $cos^{6} θ (cos^{6} θ + 3 cos^{4} θ + 3 cos^{2} θ + 1) - 1$
$= cos^{6} θ (cos^{2} θ + 1)^{3} - 1$
$= (cos^{2} θ)^{3} (sin θ + 1)^{3} - 1$
$= sin^{3} θ (sin θ + 1)^{3} - 1$
$= (sin^{2} θ + sin θ)^{3} - 1 = 1 - 1 = 0$