Choose the correct answer:
$1 - 2 {sin}^{2} 30^{\circ} =$ ______.

c o s e c 60^{\circ}

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t a n 60^{\circ}

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s i n 60^{\circ}

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c o s 60^{\circ}

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Solution

The correct option is D $c o s 60^{\circ}$
Using ${sin}^{2} x = 1 - {cos}^{2} x$
$∴ 1 - 2 {sin}^{2} 30^{\circ}$ $= 1 - 2 \times \frac{1}{4}$
$⟹ 1 - 2 {sin}^{2} 30^{\circ} = 1 - \frac{1}{2} = \frac{1}{2}$
From the given options, $cos 60^{\circ} = \frac{1}{2}$
Therefore, $1 - 2 {sin}^{2} 30^{\circ} = cos 60^{\circ}$ .

Suggest Corrections

Similar questions

Show that:

(i) $\frac{1 - sin 60^{\circ}}{cos 60^{\circ}} = \frac{tan 60^{\circ} - 1}{tan 60^{\circ} + 1}$

(ii) $\frac{cos 30^{\circ} + sin 60^{\circ}}{1 + sin 30^{\circ} + cos 60^{\circ}} = cos 30^{\circ}$

\frac{1 - \sin 60 °}{\cos 60 °} = \frac{\tan 60 ° - 1}{\tan 60 ° + 1}

\frac{\cos 30 ° + \sin 60 °}{1 + \sin 30 ° + \cos 60 °} = \cos 30 °

\frac{2 t a n 3 0^{0}}{1 - t a n^{2} 30^{0}} =

c o s 60^{\circ}

s i n 60^{\circ}

t a n 60^{\circ}

s i n 30^{\circ}

\frac{2 t a n 3 0^{0}}{1 + t a n^{2} 30^{0}} =

s i n 60^{\circ}

c o s 60^{\circ}

t a n 60^{\circ}

s i n 30^{\circ}

\frac{cos 60^{\circ}}{1 + sin 60^{\circ}} = \frac{1}{sec 60^{\circ} + tan 60^{\circ}}

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