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

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...

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

$$B = 2$$

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.

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 $$2$$.

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 $$B=4\cos{(60^∘)}=2$$

• You might want to use x^\circ instead of x° to get $x^\circ$ instead of $x°$ but that might just be personal preference. – Peter Foreman Aug 2 '19 at 22:31
• @PeterForeman Due brevity I ommited that notation but I wonder where could be the mistake that I made?. – Chris Steinbeck Bell Aug 2 '19 at 22:36
• @PeterForeman : Do the braces have any influence? $\sin$ etc. are macros without parameter, so the grouping is semantically nice but without influence on the visual output. – Lutz Lehmann Aug 2 '19 at 22:36
• I'll write them here: $\sin(x)$ and $\sin{(x)}$ (\sin(x) and \sin{(x)}). As you can see the latter has a spacing that seems intended for use with such functions. – Peter Foreman Aug 2 '19 at 22:37
• @PeterForeman : I see. One could say that it is better balanced in the distribution of black on white. – Lutz Lehmann Aug 2 '19 at 22:40

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

$$\sqrt 3 \tan 70° = 4\sin 70°+1.$$

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.

• As the OP my request was to find a method which would be geometry free or perhaps to use double angle identities?. Although this method does prove that the answer is $2$ but where could be the mistake in my computation?. – Chris Steinbeck Bell Aug 2 '19 at 22:27
• @ChrisSteinbeckBell ops, sorry. I did not get you didnt't want geometrical approach... – dfnu Aug 2 '19 at 22:28
• @ChrisSteinbeckBell, I made an edit. Maybe you can spot the mistake there? Hope it'll be useful. – dfnu Aug 2 '19 at 22:46
• Thanks! Just at the time you were posting this I was able to find the mistake by my own in my book. I wouldn't have ever imagined to make a triangle like you pictured. An isosceles triangle seems to apply to this particular situation but I wonder if it would apply to other problems like this as well. – Chris Steinbeck Bell Aug 2 '19 at 22:56
• @ChrisSteinbeckBell I honestly don't know if you can generalize... I personally like experimenting geometrical approaches to demonstrate trigonometric identities. They look more "basic" to me. But probably they're ad hoc solutions. – dfnu Aug 2 '19 at 22:59