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 August 2nd, 2019, 02:34 PM #1 Member     Joined: Jun 2017 From: Lima, Peru Posts: 99 Thanks: 1 Math Focus: Calculus How to simplify this expression involving a tangent of 70? My situation is as follows. Can the expression from below be simplified using precalculus and plain by 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 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: $\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{4\sqrt 3 \cos (40)-4\sin(40)}$ $\frac{\sin(100)-\sin(140)}{4\sqrt 3 \cos (40)-4\sin(40)}$ Then dividing by $8$ in the denominator $\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\frac{\sqrt 3}{2} \cos (40)-\frac{1}{2}\sin(40)}$ $\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\sin(60) \cos (40)-\cos(60)\sin(40)}$ $\frac{\frac{1}{8}\left(\sin(80)-\sin(40)\right)}{\sin(20)}$ Finally using prosthaphaeresis identities: $\frac{\frac{1}{8}\left(2\cos(60)\sin(20)\right)}{ \sin(20)}$ So I end up with: $\frac{1}{8}\left(2\cos(60^{\circ})\right)=\frac{1 }{4}\left(\frac{1}{2}\right)=\frac{1}{8}$ But I'm not sure whether this is an adequate method either. Does there exist a better way to simplify it or to ease calculations? Can somebody help me with an easier and quicker procedure? Apparently, the answer according to a calculator is 2. But I'm like playing Where's Wally with where I made a mistake. I could catch up that when I divided by 4 in the denominator there was a mistake, so by continuing from there: $\frac{\frac{\sqrt 3}{2} \cos (40) + \frac{1}{2} \sin(40)-\frac{\sqrt 3}{2}\cos(80)-\frac{1}{2}\sin(80)}{\frac{1}{4}\left(\sqrt 3 \cos (40)-\sin(40)\right)}$. By continuing the computation by hand, I was able to simplify the expression to: $\displaystyle \frac{\sin 100 - \sin 140}{\frac{1}{2}\left(\sin(20)\right)}=1$ But for some reason I can't find where it is the mistake. Last edited by skipjack; August 2nd, 2019 at 06:48 PM.
 August 2nd, 2019, 06:43 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 Answered here very simply, where a geometric interpretation is also given. Similar results: √3tan(10$^\circ$) + 4sin(10$^\circ$) + 1 = 2 -√3tan(50$^\circ$) + 4sin(50$^\circ$) + 1 = 2 tan(20$^\circ$) + 4sin(20$^\circ$) = √3 tan(40$^\circ$) - 4sin(40$^\circ$) = -√3 tan(80$^\circ$) - 4sin(80$^\circ$) = √3 Thanks from topsquark
 August 2nd, 2019, 09:47 PM #3 Math Team     Joined: May 2013 From: The Astral plane Posts: 2,257 Thanks: 927 Math Focus: Wibbly wobbly timey-wimey stuff. I did find a site that talked about tan(70) in general and it recommended splitting the 70 degrees into two angles of 35 degrees. But how can you get the values for sin, cos, and tan of 35 degrees? -Dan Last edited by skipjack; August 5th, 2019 at 12:13 AM.
 August 3rd, 2019, 04:55 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 There are simple equations connecting the tan and sin of various angles that are multiples of 10$^\circ$, and these can be changed to equations that involve trigonometric functions of multiples of 5$^\circ$ by use of the various trigonometric double angle identities. However, the resulting equations are more complicated. Although tan(75$^\circ\!$) is exactly 2 + √3, there isn't such an elegant expression for tan(35$^\circ\!$). Thanks from topsquark
 August 3rd, 2019, 06:37 PM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,958 Thanks: 1146 Math Focus: Elementary mathematics and beyond \displaystyle \begin{align*}B&=\sqrt{3} \tan 70^{\circ}- 4 \sin 70^{\circ}+1 \\ &=\frac{-2\cos210\sin70}{\cos70}-4\sin70+1 \\ &=\frac{-8\cos70\cos10\cos130\sin70}{\cos70}-4\sin70+1 \\ &=-8\cos10\cos130\sin70-4\sin70+1 \\ &=-4\sin70(2\cos10\cos130+1)+1 \\ &=-4\sin70(\cos120+\cos140+1)+1 \\ &=-4\sin70\left(\frac32-2\sin^270\right)+1 \\ &=-6\sin70+8\sin^370+1 \\ &=-2(3\sin70-4\sin^370)+1 \\ &=-2\sin210+1 \\ &=2\end{align*} Thanks from topsquark
 August 5th, 2019, 12:51 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,921 Thanks: 2203 The equations of the type discussed here are all effectively special cases of the identity 2cos(3x)tan(x) + 4sin(x) ≡ 2sin(3x), where x isn't an odd multiple of 90 degrees. If x = 35$^\circ\!$, note that 2cos(105$^\circ\!$) = (1 - √3)/√2 and 2sin(105$^\circ\!$) = (1 + √3)/√2. One gets simpler results if x is a multiple of 10$^\circ\!$. Thanks from greg1313

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