Show that cos 3-sin 3- -sin 3-cos 3- -3-4sin 4-

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Standard XII

Mathematics

Domain and Range of Trigonometric Ratios

Show that c...

Question

Show that ${cos}^{3} θ sin 3 θ + {sin}^{3} θ cos 3 θ = \frac{3}{4} sin 4 θ$ .

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Solution

${cos}^{3} θ sin 3 θ + {sin}^{3} θ cos 3 θ$

using trigonometric identities and simplifying

$= (\frac{cos 3 θ + 3 cos θ}{4}) sin 3 θ + (\frac{3 sin θ - sin 3 θ}{4}) cos 3 θ$

$= \frac{1}{4} (sin 3 θ cos 3 θ + 3 cos θ sin 3 θ + 3 sin 3 θ cos θ - sin 3 θ cos 3 θ)$

$= \frac{1}{4} (3 cos θ sin 3 θ + 3 sin 3 θ cos θ) = \frac{3}{4} sin 4 θ$

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Similar questions

Q. Prove that:

\cos^{3} θ \sin 3 θ + \sin^{3} θ \cos 3 θ = \frac{3}{4} \sin 3 θ

Prove that following identities:

$c o s^{3} θ s i n 3 θ + s i n^{3} θ c o s 3 θ = \frac{3}{4} s i n 4 θ$

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {sin}^{2} θ & s i n^{2} θ c o s^{2} θ & 1 + c o s^{2} θ & c o s^{2} θ 4 sin 4 θ & 4 sin 4 θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0

, then

sin 4 θ

equals to.

If $∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + s i n^{2} θ & c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & 1 + c o s^{2} θ & 4 s i n 4 θ s i n^{2} θ & c o s^{2} θ & 1 + 4 s i n 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$ such that $0 \leq θ \leq \frac{π}{2}$ then $θ$ is

Q. Solve:

∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 + {sin}^{2} θ & {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & 1 + {cos}^{2} θ & 4 sin 4 θ {sin}^{2} θ & {cos}^{2} θ & 1 + 4 sin 4 θ \end{matrix} ∣ ∣ ∣ ∣ ∣

= 0

; where,

0 < θ < \frac{π}{2}

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Show that cos3θsin3θ+sin3θcos3θ=34sin4θ.

cos3θsin3θ+sin3θcos3θusing trigonometric identities and simplifying=(cos3θ+3cosθ4)sin3θ+(3sinθ−sin3θ4)cos3θ=14(sin3θcos3θ+3cosθsin3θ+3sin3θcosθ−sin3θcos3θ)=14(3cosθsin3θ+3sin3θcosθ)=34sin4θ

Show that ${cos}^{3} θ sin 3 θ + {sin}^{3} θ cos 3 θ = \frac{3}{4} sin 4 θ$ .