If -cos 2- then find the value of sin 4-2 sin 3-sin 2-

Sec θ =cos^2 θ 1/cos θ=cos^2 θ cos^3 θ =1 cost^3 θ =√(1) cosine volume can't be imaginary ⇒ cos θ =1 If cos θ =1, sin^2 θ=1 cos^2 θ =1 1 sin^2 θ =0 ⇒ sin θ ...

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Byju's Answer

Standard XII

Mathematics

Higher Order Derivatives

If θ=cos 2θ t...

Question

If $s e c θ = c o s^{2} θ$ then find the value of $s i n^{4} θ + 2 s i n^{3} θ + s i n^{2} θ$

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Solution

$s e c θ = c o s^{2} θ \frac{1}{c o s θ} = c o s^{2} θ c o s^{3} θ = 1 c o s t^{3} θ = \sqrt[3]{1}$
cosine volume can't be imaginary
$\Rightarrow c o s θ = 1$
If $c o s θ = 1, s i n^{2} θ = 1 - c o s^{2} θ$
=1 -1
$s i n^{2} θ = 0 \Rightarrow s i n θ = 0 \to (1) ∴ s i n^{4} θ + 2 s i n^{3} θ + s i n^{2} θ = (s i n^{2} θ + s i n θ)^{2} = (0 + 0)^{2} = 0$

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If secθ=cos2θ then find the value of sin4θ+2sin3θ+sin2θ

secθ=cos2θ1cosθ=cos2θcos3θ=1cost3θ=3√1 cosine volume can't be imaginary ⇒cosθ=1 If cosθ=1,sin2θ=1−cos2θ =1 -1 sin2θ=0⇒sinθ=0→(1)∴sin4θ+2sin3θ+sin2θ=(sin2θ+sinθ)2=(0+0)2=0

If $s e c θ = c o s^{2} θ$ then find the value of $s i n^{4} θ + 2 s i n^{3} θ + s i n^{2} θ$