Haven

2021-10-25

How do you find exact value of $\mathrm{tan}\left(\frac{\pi }{4}\right)$?

Macsen Nixon

Given: $\mathrm{tan}\left(\frac{\pi }{4}\right)$
The cousine is defined as the opposite leg divided by the hypotenuse of a rectangular triangle, while the sine is defined as the adjacent leg divided by the hypotenuse.
$\mathrm{sin}\left(\frac{\pi }{4}\right)=\frac{\text{adjacent leg}}{\text{hypotenuse}}$
$\mathrm{cos}\left(\frac{\pi }{4}\right)=\frac{\text{opposite leg}}{\text{hypotenuse}}$
By the Pythagorean theorem ${a}^{2}+{b}^{2}={c}^{2}$ where a (adjacent) and b (opposite) are the legs and c is the hypotenuse c. This then implies that $c=\sqrt{{a}^{2}+{b}^{2}}$ for the hypotenuse c.
$\mathrm{sin}\left(\frac{\pi }{4}\right)=\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}$
$\mathrm{cos}\left(\frac{\pi }{4}\right)=\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}$
However, if one of the angles of the triangle is ${45}^{\circ }$ or $\frac{\pi }{4}$, then the triangle is isosceles and thus the two legs have the same length a=b
$\mathrm{cos}\left(\frac{\pi }{4}\right)=\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}$
$=\frac{b}{\sqrt{{b}^{2}+{b}^{2}}}$
$=\frac{b}{\sqrt{2{b}^{2}}}$
$=\frac{b}{\sqrt{2}\sqrt{{b}^{2}}}$
$=\frac{b}{\sqrt{2}b}$
$=\frac{1}{\sqrt{2}}$
$=\frac{1\cdot \sqrt{2}}{\sqrt{2}\cdot \sqrt{2}}$
$=\frac{\sqrt{2}}{{\left(\sqrt{2}\right)}^{2}}$
$=\frac{\sqrt{2}}{2}$
$\mathrm{sin}\left(\frac{\pi }{4}\right)=\frac{a}{\sqrt{{a}^{2}+{b}^{2}}}=\frac{b}{\sqrt{{a}^{2}+{b}^{2}}}=\mathrm{cos}\left(\frac{\pi }{4}\right)$
The tangent is the sine divided by the cousine:
$\mathrm{tan}\left(\frac{\pi }{4}\right)=\frac{\mathrm{sin}\left(\frac{\pi }{4}\right)}{\mathrm{cos}\left(\frac{\pi }{4}\right)}=\frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}}=1$
Result: 1

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