Trigo 1

Geometry Level 3

Compute 5 sin ( 1 5 ) + 5 tan ( 1 0 ) \large5\sin(15^\circ) + 5\tan(10^\circ)

Give your answer to 3 decimal places.


The answer is 2.175.

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
Mar 23, 2019

It appears that there is no algebraic solution to this so I am solving it numerically.

From cos 2 θ = 1 2 sin 2 θ \cos 2\theta = 1 - 2\sin^2 \theta :

1 2 sin 2 1 5 = cos 3 0 sin 1 5 = 1 cos 3 0 2 = 2 3 4 = 1 2 4 2 3 2 = 1 2 ( 3 1 ) 2 2 = 3 1 2 2 0.258819045 \begin{aligned} 1-2\sin^2 15^\circ & = \cos 30^\circ \\ \implies \sin 15^\circ & = \sqrt{\frac {1- \cos 30^\circ}2} = \sqrt {\frac {2-\sqrt 3}4} = \frac 12 \sqrt{\frac {4-2\sqrt 3}2} = \frac 12 \sqrt{\frac {(\sqrt 3-1)^2}2} = \frac {\sqrt 3-1}{2\sqrt 2} \approx 0.258819045 \end{aligned}

From tan 3 θ = 3 tan θ tan 3 θ 1 3 tan 2 θ \tan 3 \theta = \dfrac {3\tan \theta - \tan^3 \theta}{1-3\tan^2 \theta} :

3 tan 1 0 tan 3 1 0 1 3 tan 2 1 0 = 1 3 3 tan 3 1 0 3 tan 2 1 0 3 3 tan 1 0 + 1 = 0 \begin{aligned} \frac {3\tan 10^\circ - \tan^3 10^\circ}{1-3\tan^2 10^\circ} & = \frac 1{\sqrt 3} \\ \sqrt 3 \tan^3 10^\circ - 3 \tan^2 10^\circ - 3\sqrt 3 \tan 10^\circ + 1 & = 0 \end{aligned}

Using Newton Raphson method , we get tan 1 0 0.176326981 \tan 10^\circ \approx 0.176326981 .

Therefore, 5 sin 1 5 + 5 tan 1 0 5 ( 0.258819045 + 0.176326981 ) 2.176 5\sin 15^\circ + 5\tan 10^\circ \approx 5(0.258819045 + 0.176326981) \approx \boxed{2.176} .

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...