Prove that cos 2 48-sin 2 12-5-1-8

The correct option is We have LHS = (cos^2 48^∘ sin^2 12^∘) =1/2 (2cos^2 48^∘ 2sin^2 12^∘) =1/2 (1+cos 96^∘ ) (1 cos 24^∘ ) [ ∵ 2cos^2θ = ( 1+cos 2θ ) an ...

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Standard X

Mathematics

Trigonometric Ratios of Complementary Angles

Prove that co...

Question

Prove that $({cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ}) = \frac{(\sqrt{5} + 1)}{8}$

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Solution

We have

LHS = $({cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ})$

$= \frac{1}{2} (2 {cos}^{2} 48^{\circ} - 2 {sin}^{2} 12^{\circ})$

= $\frac{1}{2} (1 + cos 96^{\circ}) - (1 - cos 24^{\circ})$

$[∵ 2 {cos}^{2} θ = (1 + cos 2 θ) and 2 {sin}^{2} θ = (1 - cos 2 θ)]$

= $\frac{1}{2} (cos 96^{\circ} + cos 24^{\circ})$

= $\frac{1}{2} [2 cos (\frac{96^{\circ} + 24^{\circ}}{2}) cos (\frac{96^{\circ} - 24^{\circ}}{2})]$

= $\frac{1}{2} \times 2 cos 60^{\circ} 36^{\circ} = \frac{1}{2} \times \frac{(\sqrt{5} + 1)}{4} [∵ cos 36^{\circ} = \frac{(\sqrt{5} + 1)}{4}]$

= $\frac{(\sqrt{5} + 1)}{8} = RHS$

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Similar questions

Q. Show that:

{cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ} = \frac{\sqrt{5} + 1}{8}

$c o s^{2} 48^{\circ} - s i n^{2} 12^{\circ} =$

Q. Solve

c o s^{2} 48^{\circ} - s i n^{2} 12^{\circ}

({cos}^{2} 66^{\circ} - {sin}^{2} 6^{\circ}) ({cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ})

equals -

Q. The value of

\cos^{2} 48 ° - \sin^{2} 12 °

is

(a)

\frac{\sqrt{5} + 1}{8}

(b)

\frac{\sqrt{5} - 1}{8}

(c)

\frac{\sqrt{5} + 1}{5}

(d)

\frac{\sqrt{5} + 1}{2 \sqrt{2}}

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Prove that (cos2 48∘−sin2 12∘)=(√5+1)8

Prove that $({cos}^{2} 48^{\circ} - {sin}^{2} 12^{\circ}) = \frac{(\sqrt{5} + 1)}{8}$