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, $\sin{45^\circ}$ (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.

$67.5^\circ$ or $22.5^\circ$

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

$75^\circ$ or $15^\circ$

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, $\cos{75^\circ}=\sin{15^\circ}=\frac{BA}{AE}=\frac{1}{\sqrt6+\sqrt2}=\frac{\sqrt6-\sqrt2}{4}$ $\sin{75^\circ}=\cos{15^\circ}=\sqrt{1-\cos^2{75^\circ}}=\frac{\sqrt6+\sqrt2}{4}$ $\tan{75^\circ}=\cot{15^\circ}=\frac{\sin{75^\circ}}{\cos{75^\circ}}=2+\sqrt3$ $\cot{75^\circ}=\tan{15^\circ}=\frac{1}{\tan{75^\circ}}=2-\sqrt3$

$18^\circ$ or $72^\circ$

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.

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

Note by Kenneth Tan
5 years, 9 months ago

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Nice piece of information 😀

- 5 years, 9 months ago

Great!

- 5 years, 9 months ago

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

- 5 years, 9 months ago

cool.amazing piece of information. :))

- 5 years, 2 months ago

Really nice👍

- 5 years ago

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

- 3 years, 1 month ago