Proof of $\sin{(36^\circ)}$ value in Trigonometric method

In fraction form, the sine of angle thirty six degrees is equal to the square root of ten minus two times the square root of five by four.

$\sin{(36^\circ)}$ $\,=\,$ $\dfrac{\sqrt{10-2\sqrt{5}}}{4}$

Actually, it is not possible to derive the sine of angle thirty six degrees value directly from the trigonometric fundamentals but it can be derived on the basis of the cos 36 degrees value by a trigonometric identity.

Express the Sine in Cosine function

As we know that the sine and cosine functions have mutual relationship in the form of a Pythagorean identity. Hence, the sine of angle thirty six degrees can be written in terms of cosine of angle thirty six degrees in mathematical form.

$\sin^2{(36^\circ)}+\cos^2{(36^\circ)}$ $\,=\,$ $1$

$\implies$ $\sin^2{(36^\circ)}$ $\,=\,$ $1-\cos^2{(36^\circ)}$

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{1-\cos^2{(36^\circ)}}$

Now, substitute the cosine of pi by five value in the above mathematical equation for calculating the sine of angle thirty six degrees value exactly.

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{1-\Bigg(\dfrac{\sqrt{5}+1}{4}\Bigg)^2}$

Simplify the Trigonometric equation

The value of sine of angle thirty six degrees is defined in irrational form and we have to simplify this expression for evaluating the exact value of sine of angle thirty six degrees in trigonometry.

The simplification process can be initiated by the power rule of a quotient.

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{1-\dfrac{(\sqrt{5}+1)^2}{4^2}}$

Inside the square root, the square of sum of two terms can be expanded for simplifying the right hand side expression of the equation.

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{1-\dfrac{(\sqrt{5})^2+(1)^2+2 \times (\sqrt{5}) \times 1}{16}}$

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{1-\dfrac{5+1+2\sqrt{5}}{16}}$

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{1-\dfrac{6+2\sqrt{5}}{16}}$

The right hand side expression of the equation can be simplified by the subtraction rule of fractions.

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{\dfrac{1 \times 16-(6+2\sqrt{5})}{16}}$

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{\dfrac{16-(6+2\sqrt{5})}{16}}$

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{\dfrac{16-6-2\sqrt{5}}{16}}$

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \sqrt{\dfrac{10-2\sqrt{5}}{16}}$

$\implies$ $\sin{(36^\circ)}$ $\,=\,$ $\pm \dfrac{\sqrt{10-2\sqrt{5}}}{4}$

Evaluate the Sine of 36 degrees value

The trigonometric procedure has given two values for sine of angle forty grades.

$(1).\,\,\,$ $\sin{(36^\circ)}$ $\,=\,$ $+\dfrac{\sqrt{10-2\sqrt{5}}}{4}$

$(2).\,\,\,$ $\sin{(36^\circ)}$ $\,=\,$ $-\dfrac{\sqrt{10-2\sqrt{5}}}{4}$

Actually, The sine function does not give two values for one angle. So, one of them is correct and remaining one is incorrect. Now, it is time to evaluate the exact value of sine of angle pi by five radian.

The angle $36^\circ$ belongs to the first quadrant in two dimensional Cartesian coordinate system. We know that the sign of sine is positive in the first quadrant.

$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(36^\circ)}$ $\,=\,$ $\dfrac{\sqrt{10-2\sqrt{5}}}{4}$

In this way, the sin of thirty six degrees value is derived in fraction form exactly in trigonometry.

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