Trig without Tears Part 3:
Trig without Tears Part 3:
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 = 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 + a² = c² = 1
1/4 + a² = 1
a² = 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
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.
(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.
Prove: tan(180° − A) = −tan A.
next: 4/Solving Triangles
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