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:

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


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

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

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

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

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:

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

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


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

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.


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 $$ B=4\cos{(60^∘)}=2 $$

  • $\begingroup$ You might want to use x^\circ instead of to get $x^\circ$ instead of $x°$ but that might just be personal preference. $\endgroup$ – Peter Foreman Aug 2 '19 at 22:31
  • $\begingroup$ @PeterForeman Due brevity I ommited that notation but I wonder where could be the mistake that I made?. $\endgroup$ – Chris Steinbeck Bell Aug 2 '19 at 22:36
  • $\begingroup$ @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. $\endgroup$ – Lutz Lehmann Aug 2 '19 at 22:36
  • $\begingroup$ 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. $\endgroup$ – Peter Foreman Aug 2 '19 at 22:37
  • $\begingroup$ @PeterForeman : I see. One could say that it is better balanced in the distribution of black on white. $\endgroup$ – Lutz Lehmann Aug 2 '19 at 22:40

Consider the isosceles triangle in the Figure below. enter image description here 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.

  • $\begingroup$ 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?. $\endgroup$ – Chris Steinbeck Bell Aug 2 '19 at 22:27
  • $\begingroup$ @ChrisSteinbeckBell ops, sorry. I did not get you didnt't want geometrical approach... $\endgroup$ – dfnu Aug 2 '19 at 22:28
  • $\begingroup$ @ChrisSteinbeckBell, I made an edit. Maybe you can spot the mistake there? Hope it'll be useful. $\endgroup$ – dfnu Aug 2 '19 at 22:46
  • 1
    $\begingroup$ 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. $\endgroup$ – Chris Steinbeck Bell Aug 2 '19 at 22:56
  • $\begingroup$ @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. $\endgroup$ – dfnu Aug 2 '19 at 22:59

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