Sometimes we want to quickly recall back the value of a certain trigonometric function for a specific angle, say, \(\sin{45^\circ}\) (or \(\cos{45^\circ}\), \(\tan{45^\circ}\)), we can easily do this by sketching geometrical figures, in this case, a right isosceles triangle, which then we can quickly recall \(\sin{45^\circ}=\frac{\sqrt{2}}{2}\) (\(\cos{45^\circ}=\frac{\sqrt{2}}{2}\), \(\tan{45^\circ}=1\)).

But for other angles, say, \(75^\circ\) (either it is \(\sin{75^\circ}\), \(\cos{75^\circ}\) or \(\tan{75^\circ}\)) is a bit trickier.

For example to find \(\sin{75^\circ}\), we have to apply the sum formula \(\sin{(\alpha+\beta)}=\sin{\alpha}\cos{\beta}+\sin{\beta}\cos{\alpha}\) to find \(\sin{75^\circ}=\frac{\sqrt6+\sqrt2}{4}\).

This can be a little time consuming, especially for more "alien-ish" angles such as \(18^\circ\).

In this note, I will be discussing the geometrical methods for finding values for certain trigonometric functions for certain angles.

First construct a right isosceles triangle \(\triangle ABC\), assume \(AB=BC=1\), then \(AC=\sqrt2\), now extend \(BC\) to \(D\) so that \(AC=CD=\sqrt2\), connect \(AD\).

Since \(AC=CD\), we know that \(\angle DAC=\angle ADC=22.5^\circ\), \(\angle DAB=\angle DAC+\angle CAB=67.5^\circ\), thus, we have now constructed our \(22.5^\circ\)-\(67.5^\circ\)-\(90^\circ\) triangle.

By Pythagoras' theorem, we could work out \(AD=\sqrt{4+2\sqrt2}\)

Now, \[\sin{67.5^\circ}=\cos{22.5^\circ}=\frac{BD}{DA}=\frac{\sqrt2+1}{\sqrt{4+2\sqrt2}}\] \[\cos{67.5^\circ}=\sin{22.5^\circ}=\frac{BA}{AD}=\frac{1}{\sqrt{4+2\sqrt2}}\] \[\tan{67.5^\circ}=\cot{22.5^\circ}=\frac{DB}{BA}=\sqrt2+1\] \[\cot{67.5^\circ}=\tan{22.5^\circ}=\frac{AB}{BD}=\frac{1}{\sqrt2+1}=\sqrt2-1\]

Again construct a right isosceles triangle \(\triangle ABC\), assume \(AB=BC=1\), then \(AC=\sqrt2\), construct \(CD\) such that \(AC=CD=\sqrt2\) and \(\angle ACD=120^\circ\), connect \(AD\). \[\therefore \angle CAD=\angle ADC=30^\circ\] \[\angle DAB=75^\circ\]

Extend \(AD\) and \(BC\) so that they meet at \(E\), then \(\angle E=\angle ECD=15^\circ\), \(CD=DE=\sqrt2\) we have now constructed our \(15^\circ\)-\(75^\circ\)-\(90^\circ\) triangle.

By law of cosines we could work out \(AD=\sqrt6\), now, \[\cos{75^\circ}=\sin{15^\circ}=\frac{BA}{AE}=\frac{1}{\sqrt6+\sqrt2}=\frac{\sqrt6-\sqrt2}{4}\] \[\sin{75^\circ}=\cos{15^\circ}=\sqrt{1-\cos^2{75^\circ}}=\frac{\sqrt6+\sqrt2}{4}\] \[\tan{75^\circ}=\cot{15^\circ}=\frac{\sin{75^\circ}}{\cos{75^\circ}}=2+\sqrt3\] \[\cot{75^\circ}=\tan{15^\circ}=\frac{1}{\tan{75^\circ}}=2-\sqrt3\]

Consider a regular pentagon \(ABCDE\), connect \(BE\) and \(AC\) so that they intersect at \(F\), let the sides of this pentagon be \(x\) and \(AF\) be \(y\), its obvious that \(AE=AB=EF=x\), \(AF=FB=y\).

In \(\triangle AEF\), by law of cosines, \[x^2=x^2+y^2-2xy\cos{72^\circ}\] \[\therefore y=2x\cos{72^\circ}\]

In \(\triangle ABE\), by law of cosines, \[(x+y)^2=x^2+x^2-2x^2\cos{108^\circ}\] \[2xy+y^2=x^2+2x^2\cos{72^\circ}\] \[4x^2\cos{72^\circ}+4x^2\cos^2{72^\circ}=x^2(1+2\cos{72^\circ})\] \[4\cos{72^\circ}+4\cos^2{72^\circ}=1+2\cos{72^\circ}\] \[4\cos^2{72^\circ}+2\cos{72^\circ}-1=0\]

Solving this we will have \[\cos{72^\circ}=\sin{18^\circ}=\frac{\sqrt5-1}{4}\]

Hence, \[\sin{72^\circ}=\cos{18^\circ}=\sqrt{1-\cos^2{72^\circ}}=\frac{\sqrt{10+2\sqrt5}}{4}\] \[\tan{72^\circ}=\cot{18^\circ}=\frac{\sin{72^\circ}}{\cos{72^\circ}}=\frac{\sqrt5-1}{\sqrt{10+2\sqrt5}}\] \[\cot{72^\circ}=\tan{18^\circ}=\frac{1}{\tan{72^\circ}}=\frac{\sqrt{10+2\sqrt5}}{\sqrt5-1}\]

That's all for now, I hope you find this helpful.

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## Comments

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TopNewestNice piece of information 😀

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Great!

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Awesome note dude.Learned a lot.Thanks for that :D

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cool.amazing piece of information. :))

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Really nice👍

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There is another also a Geometric method for trigonometric ratios of 18 degree .Please upload it .

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Very very helpful. Thank you.

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