Trig without Tears Part 3:

# Functions of Special Angles

Copyright © 1997–2020 Stan Brown, BrownMath.com

Trig without Tears Part 3:

Copyright © 1997–2020 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,

*a*² + *b*² = *c*²

But *a* = *b* and *c* = 1; therefore

2*a*² = 1

*a*² = 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 2*b* is also 1, which means that

*b* = ½

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 + *a*² = *c*² = 1

1/4 + *a*² = 1

*a*² = 3/4 ⇒ *a* = √3/2

Since *a* = sin 60°, sin 60° = √3/2.

Knowing 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° 45° 60° 90° — or 0, π/6, π/4, π/3, 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*.

If *A* + *B* = 90°, then
angles *A* and *B* are **complements** of each
other. For example, the complement of a 40° angle is
90° − 40° = 50°.

If *A* + *B* = 180°, then
angles *A* and *B* are **supplements** of each
other. For example, the supplement of a 40° angle is
180° − 40° = 140°.

You already know about trig functions of complementary angles:
sin(90° − *A*) = cos *A*,
tan(90° − *A*) = cot *A*,
sec(90° − *A*) = csc *A*,
and vice versa. Your mnemonic is the “co” in co-function and
complementary angle.

But what about supplementary angles? Is there any relation
between sin 40° and sin 140°? As a matter of fact,
there is:
**the sine of an angle equals the sine of its supplement.** And
cosines?
**the cosine of an angle equals minus the cosine of its supplement.**
You can look ahead, if you want, to
see why that’s true, or for now you can just take it as a given,
while we work some practical problems in the next part of this
book.

To get the most benefit from these problems, work them without first looking at the solutions. Refer back to the chapter text if you need to refresh your memory.

**Special advice:** Don’t be afraid to draw a picture of a 45-45-90° or 30-60-90° triangle if you need to, especially while you’re first getting used to the functions of the special angles.

**Recommendation**: Work them on paper —
it’s harder to fool yourself about whether you really
understand a problem completely.

You’ll find full solutions for all problems. Don’t just check your answers, but check your method too.

1 (a) Sketch a 45-45-90° triangle with hypotenuse of 1.
Label the size of each angle and the exact length of each side, not a
calculator approximation. (Hint: Since the two acute angles are equal,
the two short sides must be equal. That and the Theorem of Pythagoras
is enough to let you find them.)

(b) Draw a triangle with all sides equal to 1. If all sides are equal, all angles must be equal. Knowing that they add to 180°, fill in the value of each angle. Now drop a perpendicular from the top of the triangle to the middle of the opposite side. You now have two 30-60-90° triangles with hypotenuse of 1. Fill in the exact lengths of the short sides of those two triangles.

For the rest of these problems, refer to these sketches
if you need to.
**Give exact answers, not decimal approximations.**

2
Find tan 45°, cos 45°, sin 90°,
cos 30°,
sin 30°,
cos 90°.

3
Find sin(π/4), cos(π/6), tan(π/3).

4
Find each of the following angles from the clues, assuming all the
angles are between 0 and π/2 (0° and 90°) inclusive. Give
each answer in degrees and radians.
sin *A* = 0;
cos *B* = √3/2;
sin *C* = 1/2;
sin *D* = 1;
tan *E* = 1;
cos *F* = 1/2;
tan *G* = 0;
tan *H* = √3;
cos *I* = 1;
cos *J* = 0.

5
Find sec 60° and cot 30°. Hint: Remember how the
secant and cotangent are defined in terms of the
“big three” functions sine, cosine, and tangent.

6
Find sin 120°, cos 120°, and tan 120°.

7Even though you can always get the tangent of a supplementary
angle from the sine and cosine, it’s a time-saver to have a rule
for the supplement of a tangent. The last problem’s solution
suggested what that rule might be.

Prove: tan(180° − *A*) =
−tan *A*.

8
Find tan 150°.

**29 Nov 2016**:- Added the section Complements and Supplements.
- Added some practice problems for complements and supplements.
- Removed lots of redundant parentheses in equation 16, making it much easier to read.

**19 Nov 2016**: Added practice problems.**1 Nov 2016**: Updated the mathematical notation, particularly the use of italics and spaces, to conform to the standard. I used Jukka Korpela’s comprehensive Writing Mathematical Expressions (2014, Suomen E-painos Oy), ISBN 978-952-6613-25-3.- (intervening changes suppressed)
**2 Mar 1997**: New document.

next: 4/Solving Triangles

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