Geometrical Approach for Finding Values for Certain Trigonometric Functions

Sometimes we want to quickly recall back the value of a certain trigonometric function for a specific angle, say, sin45\sin{45^\circ} (or cos45\cos{45^\circ}, tan45\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 sin45=22\sin{45^\circ}=\frac{\sqrt{2}}{2} (cos45=22\cos{45^\circ}=\frac{\sqrt{2}}{2}, tan45=1\tan{45^\circ}=1).

But for other angles, say, 7575^\circ (either it is sin75\sin{75^\circ}, cos75\cos{75^\circ} or tan75\tan{75^\circ}) is a bit trickier.

For example to find sin75\sin{75^\circ}, we have to apply the sum formula sin(α+β)=sinαcosβ+sinβcosα\sin{(\alpha+\beta)}=\sin{\alpha}\cos{\beta}+\sin{\beta}\cos{\alpha} to find sin75=6+24\sin{75^\circ}=\frac{\sqrt6+\sqrt2}{4}.

This can be a little time consuming, especially for more "alien-ish" angles such as 1818^\circ.

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

67.567.5^\circ or 22.522.5^\circ

First construct a right isosceles triangle ABC\triangle ABC, assume AB=BC=1AB=BC=1, then AC=2AC=\sqrt2, now extend BCBC to DD so that AC=CD=2AC=CD=\sqrt2, connect ADAD.

Since AC=CDAC=CD, we know that DAC=ADC=22.5\angle DAC=\angle ADC=22.5^\circ, DAB=DAC+CAB=67.5\angle DAB=\angle DAC+\angle CAB=67.5^\circ, thus, we have now constructed our 22.522.5^\circ-67.567.5^\circ-9090^\circ triangle.

By Pythagoras' theorem, we could work out AD=4+22AD=\sqrt{4+2\sqrt2}

Now, sin67.5=cos22.5=BDDA=2+14+22\sin{67.5^\circ}=\cos{22.5^\circ}=\frac{BD}{DA}=\frac{\sqrt2+1}{\sqrt{4+2\sqrt2}} cos67.5=sin22.5=BAAD=14+22\cos{67.5^\circ}=\sin{22.5^\circ}=\frac{BA}{AD}=\frac{1}{\sqrt{4+2\sqrt2}} tan67.5=cot22.5=DBBA=2+1\tan{67.5^\circ}=\cot{22.5^\circ}=\frac{DB}{BA}=\sqrt2+1 cot67.5=tan22.5=ABBD=12+1=21\cot{67.5^\circ}=\tan{22.5^\circ}=\frac{AB}{BD}=\frac{1}{\sqrt2+1}=\sqrt2-1

7575^\circ or 1515^\circ

Again construct a right isosceles triangle ABC\triangle ABC, assume AB=BC=1AB=BC=1, then AC=2AC=\sqrt2, construct CDCD such that AC=CD=2AC=CD=\sqrt2 and ACD=120\angle ACD=120^\circ, connect ADAD. CAD=ADC=30\therefore \angle CAD=\angle ADC=30^\circ DAB=75\angle DAB=75^\circ

Extend ADAD and BCBC so that they meet at EE, then E=ECD=15\angle E=\angle ECD=15^\circ, CD=DE=2CD=DE=\sqrt2 we have now constructed our 1515^\circ-7575^\circ-9090^\circ triangle.

By law of cosines we could work out AD=6AD=\sqrt6, now, cos75=sin15=BAAE=16+2=624\cos{75^\circ}=\sin{15^\circ}=\frac{BA}{AE}=\frac{1}{\sqrt6+\sqrt2}=\frac{\sqrt6-\sqrt2}{4} sin75=cos15=1cos275=6+24\sin{75^\circ}=\cos{15^\circ}=\sqrt{1-\cos^2{75^\circ}}=\frac{\sqrt6+\sqrt2}{4} tan75=cot15=sin75cos75=2+3\tan{75^\circ}=\cot{15^\circ}=\frac{\sin{75^\circ}}{\cos{75^\circ}}=2+\sqrt3 cot75=tan15=1tan75=23\cot{75^\circ}=\tan{15^\circ}=\frac{1}{\tan{75^\circ}}=2-\sqrt3

1818^\circ or 7272^\circ

Consider a regular pentagon ABCDEABCDE, connect BEBE and ACAC so that they intersect at FF, let the sides of this pentagon be xx and AFAF be yy, its obvious that AE=AB=EF=xAE=AB=EF=x, AF=FB=yAF=FB=y.

In AEF\triangle AEF, by law of cosines, x2=x2+y22xycos72x^2=x^2+y^2-2xy\cos{72^\circ} y=2xcos72\therefore y=2x\cos{72^\circ}

In ABE\triangle ABE, by law of cosines, (x+y)2=x2+x22x2cos108(x+y)^2=x^2+x^2-2x^2\cos{108^\circ} 2xy+y2=x2+2x2cos722xy+y^2=x^2+2x^2\cos{72^\circ} 4x2cos72+4x2cos272=x2(1+2cos72)4x^2\cos{72^\circ}+4x^2\cos^2{72^\circ}=x^2(1+2\cos{72^\circ}) 4cos72+4cos272=1+2cos724\cos{72^\circ}+4\cos^2{72^\circ}=1+2\cos{72^\circ} 4cos272+2cos721=04\cos^2{72^\circ}+2\cos{72^\circ}-1=0

Solving this we will have cos72=sin18=514\cos{72^\circ}=\sin{18^\circ}=\frac{\sqrt5-1}{4}

Hence, sin72=cos18=1cos272=10+254\sin{72^\circ}=\cos{18^\circ}=\sqrt{1-\cos^2{72^\circ}}=\frac{\sqrt{10+2\sqrt5}}{4} tan72=cot18=sin72cos72=5110+25\tan{72^\circ}=\cot{18^\circ}=\frac{\sin{72^\circ}}{\cos{72^\circ}}=\frac{\sqrt5-1}{\sqrt{10+2\sqrt5}} cot72=tan18=1tan72=10+2551\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.

This is one part of 1+1 is not = to 3.

Note by Kenneth Tan
5 years, 11 months ago

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

Dhiraj Kushwaha - 3 years, 2 months ago

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Really nice👍

will jain - 5 years, 1 month ago

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cool.amazing piece of information. :))

Satyabrata Dash - 5 years, 3 months ago

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Very very helpful. Thank you.

Niranjan Khanderia - 5 years, 4 months ago

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

Athiyaman Nallathambi - 5 years, 11 months ago

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Rohit Udaiwal - 5 years, 11 months ago

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Nice piece of information 😀

Swapnil Das - 5 years, 11 months ago

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