$3 sin 40^{o} - 4 {sin}^{3} 40^{o} = ?$

\frac{\sqrt{3}}{2}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

3 / 2

No worries! We‘ve got your back. Try BYJU‘S free classes today!

3 \sqrt{2}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

n o n e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $\frac{\sqrt{3}}{2}$
We have,
$3 sin 40^{0} - 4 {sin}^{3} 40^{0}$

We know that
$sin 3 x = 3 sin x - 4 {sin}^{3} x$

Therefore,
$\Rightarrow sin (3 \times 40^{0})$
$= sin (120^{0})$
$= \frac{\sqrt{3}}{2}$

Hence, this is the answer.

Suggest Corrections

Similar questions

$\frac{1}{(3 + 2 \sqrt{2})} = ?$

(a) $\frac{3 - 2 \sqrt{2}}{17}$

(b) $\frac{(3 - 2 \sqrt{2})}{13}$

(d)None of these

\frac{1}{(\sqrt{4} - \sqrt{3})} = ?

(2 + \sqrt{3})

(2 - \sqrt{3})

N O_{3}^{-}, C O_{3}^{2 -}, C l O_{3}^{-}, S O_{2}

Which among the following are isoelectronic $N O_{3}^{-}, C O_{3}^{2 -}, C l O_{3}^{-}, S O_{2}$

$(12 m^{2} - 27) = ? (a) (2 m - 3) (3 m - 9) (b) 3 (2 m - 9) (3 m - 1) (c) 3 (2 m - 3) (2 m + 3) (d) n o n e o f t h e s e$

Join BYJU'S Learning Program