Trig without Tears Part 3:

# Functions of Special Angles

revised 4 Apr 2016

Copyright © 1997–2016 Stan Brown, BrownMath.com

Trig without Tears Part 3:

revised 4 Apr 2016

Copyright © 1997–2016 Stan Brown, BrownMath.com

**Summary:**
**You need to know** the function values of certain special
angles, namely
**30° (π/6), 45° (π/4), and 60° (π/3)**.
You also need to be able to go backward and know
what angle has a sine of ½ or a tangent of −√3.
While it’s easy to **work them out as you go** (using easy right
triangles), you really need to **memorize them** because you’ll
use them so often that deriving them or looking them up every time
would really slow you down.

Look at this 45-45-90° triangle, which means sides a and b are equal. By the Pythagorean theorem,

a² + b² = c²

But a = b and c = 1; therefore

2a² = 1

a² = 1/2

a = 1/√2 = (√2)/2

Since a = *sin* 45°,

*sin* 45° = (√2)/2

Also, b = *cos* 45° and b = a;
therefore

*cos* 45° = (√2)/2

Use the definition of *tan* A, equation 3 or equation 4:

*tan* 45° = a/b = 1

(14) *sin* 45° = *cos* 45° = (√2)/2

*tan* 45° = 1

Now look at this diagram. I’ve drawn two 30-60-90° triangles back to back, so that the two 30° angles are next to each other. Since 2×30° = 60°, the big triangle is a 60-60-60° equilateral triangle. Each of the small triangles has hypotenuse 1, so the length 2b is also 1, which means that

b = ½2s

But b also equals *cos* 60°, and therefore

*cos* 60° = ½

You can find a, which is *sin* 60°,
by using the Pythagorean theorem:

(½)² + a² = c² = 1

1/4 + a² = 1

a² = 3/4 ⇒ a = (√3)/2

Since a = *sin* 60°, *sin* 60° = (√3)/2.

Since you know the sine and cosine of 60°, you can easily use the cofunction identities (equation 2) to get the cosine and sine of 30°:

*cos* 30° = *sin*(90°−30°) =
*sin* 60° = (√3)/2

*sin* 30° = *cos*(90°−30°) =
*cos* 60° = 1/2

As before, use the definition of the tangent to find the tangents of 30° and 60° from the sines and cosines:

*tan* 30° = *sin* 30° / *cos* 30°

*tan* 30° = (1/2) / ((√3)/2)

*tan* 30° = 1 / √3 = (√3)/3

and

*tan* 60° = *sin* 60° / *cos* 60°

*tan* 60° = ((√3)/2) / (1/2)

*tan* 60° = √3

The values of the trig functions of 30° and 60° can be summarized like this:

(15) *sin* 30° = ½, *sin* 60° = (√3)/2

*cos* 30° = (√3)/2, *cos* 60° = ½

*tan* 30° = (√3)/3, *tan* 60° = √3

Incidentally, the sines and cosines of 0, 30°, 45°, 60° and 90° display a pleasing pattern:

(16) for angle A = 0, 30° (π/6), 45° (π/4), 60° (π/3), 90° (π/2):

*sin* A = (√0)/2, (√1)/2, (√2)/2, (√3)/2, (√4)/2

*cos* A = (√4)/2, (√3)/2, (√2)/2, (√1)/2, (√0)/2

*tan* A = 0, (√3)/3, 1, √3, *undefined*

It’s not surprising that the cosine pattern is a mirror image of the
sine pattern, since *sin*(90°−A) = *cos* A.

next: 4/Functions of Any Angle

Because this textbook helps you,

please click to donate!Because this textbook helps you,

please donate at

BrownMath.com/donate.

please click to donate!Because this textbook helps you,

please donate at

BrownMath.com/donate.

Updates and new info: http://BrownMath.com/twt/