Trigonometry Formulae - Part 10

Geometry Level 3

Find the value of sin 5 0 ( 1 + 3 tan 1 0 ) \sin50^\circ\left(1+\sqrt{3}\tan10^\circ\right) .


The answer is 1.

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Chew-Seong Cheong
Jul 27, 2019

X = sin 5 0 ( 1 + 3 tan 1 0 ) = sin ( 6 0 1 0 ) × cos 1 0 + 3 sin 1 0 cos 1 0 = ( 3 2 cos 1 0 1 2 sin 1 0 ) cos 1 0 + 3 sin 1 0 cos 1 0 = 3 cos 2 1 0 + 2 sin 1 0 cos 1 0 3 sin 2 1 0 2 cos 1 0 = 3 cos 2 0 + sin 2 0 2 cos 1 0 = sin 6 0 cos 2 0 + cos 6 0 sin 2 0 cos 1 0 = sin 8 0 cos 1 0 = cos 1 0 cos 1 0 = 1 \begin{aligned} X & = \sin 50^\circ (1+\sqrt 3 \tan 10^\circ) \\ & = \sin (60^\circ - 10^\circ) \times \frac {\cos 10^\circ +\sqrt 3 \sin 10^\circ}{\cos 10^\circ} \\ & = \left(\frac {\sqrt 3}2 \cos 10^\circ - \frac 12 \sin 10^\circ \right)\frac {\cos 10^\circ +\sqrt 3 \sin 10^\circ}{\cos 10^\circ} \\ & = \frac {\sqrt 3\cos^2 10^\circ + 2\sin 10^\circ \cos 10^\circ - \sqrt 3 \sin^2 10^\circ}{2\cos 10^\circ} \\ & = \frac {\sqrt 3\cos 20^\circ + \sin 20^\circ}{2\cos 10^\circ} \\ & = \frac {\sin 60^\circ \cos 20^\circ + \cos 60^\circ \sin 20^\circ}{\cos 10^\circ} \\ & = \frac {\sin 80^\circ}{\cos 10^\circ} = \frac {\cos 10^\circ}{\cos 10^\circ} = \boxed 1 \end{aligned}

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...