Chapter 6.2, Problem 133AYU

To determine

Explain how to quickly compute the trigonometric functions of $30°,45°,60°$ .

Answer to Problem 133AYU

  $\boxed{\begin{array}{l}\sin 30°=\frac{1}{2},\cos 30°=\frac{\sqrt{3}}{2},\tan 30°=\frac{1}{\sqrt{3}},\csc 30°=2,\sec 30°=\frac{2}{\sqrt{3}},\cot 30°=\sqrt{3}\\ \sin 45°=\frac{\sqrt{2}}{2},\cos 45°=\frac{\sqrt{2}}{2}\tan 45°=1,\csc 45°=\sqrt{2},\cot 45°=1\\ \sin 60°=\frac{\sqrt{3}}{2}\cos 60°=\frac{1}{2},\tan 60°=\sqrt{3},\csc 60°=\frac{2}{\sqrt{3}},\sec 60°=2,\cot 60°=\frac{1}{\sqrt{3}}\end{array}}$

Explanation of Solution

Given information:

Write a brief paragraph that explain how to quickly compute the trigonometric functions of $30°,45°,60°$ .

Calculation:

Consider a right triangle as shown.

  

By using angle sum property of triangle, equal angle have equal side opposite to them,

So,

  $a=b\mathrm{........}\left(1\right)$

Now apply Pythagorean Theorem we get,

  $\begin{array}{l}{c}^2=a^2+b^2\\ c^2=a^2+a^2\left[\therefore a=b\right]\\ c^2=2a^2\\ c=\sqrt{2a}\mathrm{........}\left(2\right)\end{array}$

Now,

  $\begin{array}{l}\sin 45°=\frac{b}{c}\\ =\frac{a}{\sqrt{2a}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\end{array}$

  $\begin{array}{l}\cos 45°=\frac{a}{c}\\ =\frac{a}{\sqrt{2a}}=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}\end{array}$

Now reciprocate the identities,

  $\tan 45°=1,\csc 45°=\sqrt{2},\cot 45°=1$

Now consider equilateral triangle as shown,

  

  $\begin{array}{l}\angle BAD=\frac{60°}{2}\\ =30°\\ BD=\frac{BC}{2}\\ =\frac{a}{2}\end{array}$

Apply Pythagorean Theorem to $\Delta BAD$

  $\begin{array}{l}A{D}^2=a^2-{\left(\frac{a}{2}\right)}^2\\ =a^2-\frac{a^2}{4}\\ =\frac{3a^2}{4}\\ AD=\frac{\sqrt{3}a}{2}\end{array}$

  $\begin{array}{l}\sin 60°=\frac{\frac{\sqrt{3}a}{2}}{a}\\ =\frac{\sqrt{3}}{2}\\ \cos 60°=\frac{\frac{a}{2}}{a}\\ =\frac{1}{2}\end{array}$

Now by using reciprocal and quotient identity,

  $\tan 60°=\sqrt{3},\csc 60°=\frac{2}{\sqrt{3}},\sec 60°=2,\cot 60°=\frac{1}{\sqrt{3}}$

  $\begin{array}{l}\sin 30°=\frac{\frac{a}{2}}{a}\\ =\frac{1}{2}\\ \cos 30°=\frac{\frac{\sqrt{3}a}{2}}{a}\\ =\frac{\sqrt{3}}{2}\end{array}$

Now by using reciprocal and quotient identity,

  $\tan 30°=\frac{1}{\sqrt{3}},\csc 30°=2,\sec 30°=\frac{2}{\sqrt{3}},\cot 30°=\sqrt{3}$

Hence,

  $\boxed{\begin{array}{l}\sin 30°=\frac{1}{2},\cos 30°=\frac{\sqrt{3}}{2},\tan 30°=\frac{1}{\sqrt{3}},\csc 30°=2,\sec 30°=\frac{2}{\sqrt{3}},\cot 30°=\sqrt{3}\\ \sin 45°=\frac{\sqrt{2}}{2},\cos 45°=\frac{\sqrt{2}}{2}\tan 45°=1,\csc 45°=\sqrt{2},\cot 45°=1\\ \sin 60°=\frac{\sqrt{3}}{2}\cos 60°=\frac{1}{2},\tan 60°=\sqrt{3},\csc 60°=\frac{2}{\sqrt{3}},\sec 60°=2,\cot 60°=\frac{1}{\sqrt{3}}\end{array}}$