If x = cos 10^circ cos 20^circ cos 40^circ , | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given: $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ$Multiply and divide by $2 \sin 10^\circ$:$x = \frac{2 \sin 10^\circ \cos 10^\circ \cos 20^\c

Vedclass · Accepted Answer

$\frac{1}{8}\cot 10^\circ $

Vedclass · Answer

(B) Given: $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ$Multiply and divide by $2 \sin 10^\circ$:$x = \frac{2 \sin 10^\circ \cos 10^\circ \cos 20^\circ \cos 40^\circ}{2 \sin 10^\circ}$Using $2 \sin A \cos A = \sin 2A$:$x = \frac{\sin 20^\circ \cos 20^\circ \cos 40^\circ}{2 \sin 10^\circ}$Multiply and divide by $2$:$x = \frac{2 \sin 20^\circ \cos 20^\circ \cos 40^\circ}{4 \sin 10^\circ} = \frac{\sin 40^\circ \cos 40^\circ}{4 \sin 10^\circ}$Multiply and divide by $2$ again:$x = \frac{2 \sin 40^\circ \cos 40^\circ}{8 \sin 10^\circ} = \frac{\sin 80^\circ}{8 \sin 10^\circ}$Since $\sin 80^\circ = \cos 10^\circ$:$x = \frac{\cos 10^\circ}{8 \sin 10^\circ} = \frac{1}{8} \cot 10^\circ$.

If $x = \cos 10^\circ \cos 20^\circ \cos 40^\circ ,$ then the value of $x$ is

Similar Questions

The value of $\cos 24^{\circ} + \cos 5^{\circ} + \cos 300^{\circ} + \cos 175^{\circ} + \cos 204^{\circ}$ is

If the sides of a right-angled triangle are $\{\cos2\alpha + \cos2\beta + 2\cos(\alpha + \beta )\}$ and $\{\sin2\alpha + \sin2\beta + 2\sin(\alpha + \beta )\}$,then the length of the hypotenuse is:

$\tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ = $

$\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ = $ (in $/16$)

$\sin (\beta + \gamma - \alpha ) + \sin (\gamma + \alpha - \beta ) + \sin (\alpha + \beta - \gamma ) - \sin (\alpha + \beta + \gamma ) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module