How to simplify $B=\sqrt{3} \tan 70^{\circ}- 4 \sin 70^{\circ}+1$?

Question

My situation is as follows. Can the expression from below be simplified using the concept of precalculus (i.e. via hand calculation) without requiring a calculator?

$$B=\sqrt{3} \tan 70^{\circ}- 4 \sin 70^{\circ}+1$$

What I attempted to do was to split the functions in a sum of $30^{\circ}+40^{\circ}$ since the trigonometric expressions for $30^{\circ}$ is 'known'.

By going into that route, I went through this as shown below:

$\sqrt{3} \tan\left(30+40\right)-4\sin\left(30+40\right)+1$

$\sqrt{3}\left(\frac{\tan(30)+\tan(40)}{1-\tan(30)\tan(40)}\right)-4(\sin(30)\cos(40)+\cos(30)\sin(40)+1$

$\sqrt{3}\left(\frac{\frac{1}{\sqrt 3}+\tan(40)}{1-\frac{1}{\sqrt 3}\tan(40)}\right)-4\left(\frac{1}{2}\cos(40)+\frac{\sqrt 3}{2}\sin(40)\right)+1$

$\sqrt{3}\left(\frac{1+\sqrt 3\tan(40)}{\sqrt 3-\tan(40)}\right)-2\cos(40)-2\sqrt {3} \sin(40)+1$

$\frac{\sqrt 3 + 3\frac{\sin(40)}{\cos(40)}}{\sqrt 3-\frac{\sin(40)}{\cos(40)}}-2\cos(40)-2\sqrt {3} \sin(40)+1$

$\frac{\sqrt 3 \cos (40) + 3 \sin(40)}{\sqrt 3 \cos (40)-\sin(40)}-2\cos(40)-2\sqrt {3} \sin(40)+1$

Then multiplying by $\sqrt 3 \cos (40)-\sin(40)$

$\frac{\sqrt 3 \cos (40) + 3 \sin(40)-2\sqrt 3\cos^2(40)+2\sin(40)\cos(40)-6\sin(40)\cos(40)+2\sqrt{3}\sin^2(40)+\sqrt 3 \cos (40)-\sin(40)}{\sqrt 3 \cos (40)-\sin(40)}$

$\frac{2\sqrt 3 \cos (40) + 2 \sin(40)-2\sqrt 3\cos(80)-2\sin(80)}{\sqrt 3 \cos (40)-\sin(40)}$

Now dividing by $4$ on the numerator:

$\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{(\frac{1}{4})\sqrt 3 \cos (40)-(\frac{1}{4})\sin(40)}$

$\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{(\frac{1}{4})\sqrt 3 \cos (40)-(\frac{1}{4})\sin(40)}$

$\frac{\sin 60 \cos (40) + \cos 60 \sin(40)-\sin 60\cos(80)-\cos 60\sin(80)}{(\frac{1}{2})\left((\frac{1}{2})\sqrt 3 \cos (40)-(\frac{1}{2})\sin(40)\right)}$

$\frac{\sin 60 \cos (40) + \cos 60 \sin(40)-\sin 60\cos(80)-\cos 60\sin(80)}{(\frac{1}{2})\left(\sin 60 \cos (40)-\cos 60\sin(40)\right)}$

$\frac{\sin 100-\sin 140}{(\frac{1}{2})\left(\sin 20\right)}$

Using prosthaphaeresis identities:

$\frac{\sin 80-\sin 40}{(\frac{1}{2})\left(\sin 20\right)}$

$\frac{2\cos 60 \sin 20 }{(\frac{1}{2})\left(\sin 20\right)}$

Finally...

$\frac{2\left(\frac{1}{2}\right) \sin 20 }{(\frac{1}{2})\left(\sin 20\right)}$

Therefore, the answer becomes:

$$B = 2$$

So far this is the answer which I got and it seems to check with what the calculator says it is.

But I'm not sure if this is an adequate method neither does it exist in a way to better simplify it or to ease calculations. Can somebody help me with an easier and quicker procedure? If possible, without geometry.

Answer 1

Your error: Instead of canceling a factor $4$, that is, equally dividing by $4$ in numerator and denominator, you multiplied by $4$ in the denominator. This is twice the error, first missing to divide and then multiplying, and results in a wrong additional factor $\frac1{4^2}$ for the fraction from then on. And indeed $16\cdot\frac18$ gives the correct result $2$.

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Using trigonometric identities and the trig. values at $30°$ one gets in a shorter calculation leaving the denominator unchanged: \begin{align} \frac12B\cos{(70^∘)}&=\cos{(30^∘)}\sin{(70^∘)}+\sin{(30^∘)}\cos{(70^∘)}-2\sin{(70^∘)}\cos{(70^∘)} \\ &=\sin{(100^∘)}-\sin{(140^∘)} \\ &=\sin{(80^∘)}-\sin{(40^∘)} \\ &=2\cos{(60^∘)}\sin{(20^∘)} \\ &=2\cos{(60^∘)}\cos{(70^∘)} \end{align} so that in the end $$ B=4\cos{(60^∘)}=2 $$

Answer 2

Consider the isosceles triangle in the Figure below. Let $\overline{AB} = 2\sqrt 3$ and $\angle CAB = \angle CBA = 70°$. $CH$ is the altitude and $\angle DAB =30°$.

Then you have $\overline{AD} = 2$ and $\overline{DH} = 1$.

Draw from $D$ the line parallel to $AB$ that meets $AC$ in $E$. Also take $F$ on $CE$ so that $\angle FDE = 70°$.

$\triangle ADF$ is isosceles, thus $\overline{DF} = 2$.

$\triangle DEF$ is isosceles so $\overline{EF} = 2$.

$\triangle DFC$ is isosceles so $\overline{FC} = 2$.

$\overline{CE} = 4$, and then $\overline{CD} = 4\sin 70°$.

Your expression comes from the relationship

$$\overline{CH} = \overline{CD}+\overline{DH},$$

that is

$$\sqrt 3 \tan 70° = 4\sin 70°+1.$$

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I'll review your steps from here (up to this step everyhing is correct):

\begin{eqnarray} B&=& \frac{2\sqrt 3 \cos (40) + 2 \sin(40)-2\sqrt 3\cos(80)-2\sin(80)}{\sqrt 3 \cos (40)-\sin(40)}=\\ &=& \frac{4\left[\frac{\sqrt 3}2 \cos (40) + \frac12 \sin(40)-\frac{\sqrt 3}2\cos(80)-\frac12\sin(80)\right]}{2 \left[\frac{\sqrt 3}2 \cos (40)-\frac12\sin(40)\right]}=\\ &=&2\frac{\sin (100) -\sin (140)}{\sin(20)}. \end{eqnarray} Then everything is correct again, I think.

Answer 3

Two years late but here's another solution that requires a little less computation. $$B=\sqrt{3}\tan(70^{\circ})-4\sin(70^{\circ})+1\\ =\frac{\sqrt{3}\sin(70^{\circ})}{\cos(70^{\circ})}-4\sin(70^{\circ})+1\\ =2\sec(70^{\circ})\left(\frac{\sqrt{3}}{2}\sin(70^{\circ})-2\sin(70^{\circ})\cos(70^{\circ})\right)+1\\ =2\sec(70^{\circ})\Big(\sin(60^{\circ})\sin(70^{\circ})+\sin(270^{\circ})\sin(140^{\circ})\Big)+1$$

Using the product-to-sum formula for sine gives

$$B=2\sec(70^{\circ})\left(\frac{1}{2}\right)\Big(\cos(10^{\circ})-\cos(130^{\circ})+\cos(130^{\circ})-\cos(410^{\circ})\Big)+1\\ =\sec(70^{\circ})\Big(\cos(10^{\circ})-\cos(410^{\circ})\Big)+1$$

Sum-to-product formula for cosine gives

$$B=-2\sec(70^{\circ})\sin(210^{\circ})\sin(-200^{\circ})+1\\ =-\frac{\sin(200^{\circ})}{\cos(70^{\circ})}+1\\ =-\frac{(-\sin(20^{\circ}))}{\sin(20^{\circ})}+1=2$$

recalling that $\sin(180^{\circ}+\theta)=-\sin(\theta)$ and $\cos(90^{\circ}-\theta)=\sin(\theta)$