Prove thatsin-2 sin 3-2 cos 3-cos-tan-4 MARKS -

Formula: 1 Mark Concept: 1 Mark Application: 2 Marks sin θ 2 sin^3 θ/2 cos^3 θ cos θ =sin θ(1 2 sin^2 θ)/cos θ(2 cos^2 θ 1) =tan θ(sin^2 θ+cos^2 θ 2 sin^2 θ) ...

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Standard X

Mathematics

Trigonometric Identity- 1

Prove thatsin...

Question

Prove that

$\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ} = t a n θ$ [4 MARKS]

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Solution

Formula: 1 Mark
Concept: 1 Mark
Application: 2 Marks

$\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ}$

$= \frac{s i n θ (1 - 2 s i n^{2} θ)}{c o s θ (2 c o s^{2} θ - 1)}$

$= \frac{t a n θ (s i n^{2} θ + c o s^{2} θ - 2 s i n^{2} θ)}{(2 c o s^{2} θ - s i n^{2} θ - c o s^{2} θ)}$

$= \frac{t a n θ (c o s^{2} θ - s i n^{2} θ)}{(c o s^{2} θ - s i n^{2} θ)}$

$= t a n θ$

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Q. Prove

\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ} = tan θ

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\frac{sin θ - 2 {sin}^{3} θ}{2 {cos}^{3} θ - cos θ} = tan θ

Solve the following:

\frac{s i n Θ - 2 s i n^{3} Θ}{2 c o s^{3} Θ - c o s Θ} = t a n Θ

Q. Question 5 (vii)
Prove the following identities, where the angles involved are acute angles for which the expressions are defined.
(vii)

\frac{(s i n θ - 2 s i n^{3} θ)}{(2 c o s^{3} θ - c o s θ)} = t a n θ

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Prove that sin θ−2 sin3 θ2 cos3 θ−cos θ=tan θ [4 MARKS]

Formula: 1 Mark Concept: 1 Mark Application: 2 Marks sin θ−2 sin3 θ2 cos3 θ−cos θ =sin θ(1−2 sin2 θ)cos θ(2 cos2 θ−1) =tan θ(sin2 θ+cos2 θ−2 sin2 θ)(2 cos2 θ−sin2 θ−cos2 θ) =tan θ(cos2 θ−sin2 θ)(cos2 θ−sin2 θ) =tan θ

Prove that

$\frac{s i n θ - 2 s i n^{3} θ}{2 c o s^{3} θ - c o s θ} = t a n θ$ [4 MARKS]