Question

Evaluate cot 45° without using a calculator by using ratios in a reference triangle

Answer 1

Answer on Question #60579 – Math – Trigonometry

Question

Evaluate cot $45{}^{\circ}$ without using a calculator by using ratios in a reference triangle.

Solution

Given

$$\alpha = 45{}^{\circ} \ ,$$ $$\beta = 45{}^{\circ} \ ,$$ $$\gamma = 90{}^{\circ} \ ,$$

triangle $\Delta ABC$ is isosceles, because two angles $\alpha, \beta$ of the triangle being equal.

Then $a = b$, because triangle $\Delta ABC$ is isosceles. Triangle $\Delta ABC$ is right, because $\gamma = 90{}^{\circ}$.

By Pythagorean theorem,

$$c = \sqrt{a^2 + b^2} = \sqrt{a^2 + a^2} = a\sqrt{1 + 1} = a\sqrt{2}.$$

Using definitions of trigonometric functions

$$\sin \alpha = \frac{a}{c} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}},$$ $$\cos \alpha = \frac{b}{c} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}},$$ $$\cot \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}} = 1.$$

Given $\alpha = 45{}^{\circ}$, formula cot $45{}^{\circ} = 1$ has been proved.

**Answer**: cot $45{}^{\circ} = 1$.

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