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# The Value of Special Angles of Trigonometric Functions

Example 1

The terminal side of angle a is pi/6. The radius of the circle is one, that is, r = 1.The P ( x , y ) is the intersection of the circle and the terminal side . Find the value of x and y, sin a , cos a , and tan a.

Solution

In right triangle OPM, since angle a = pi/6 = 30o, so the length of its opposite is one-half of the hypotenuse, that is, PM = r/2 , since r = 1 (given), so PM = 1/2. Since sin a = PM/r = PM/1 = PM, so PM is called the line of sin a. Since cos a = OM/r = OM/1 = OM, so OM is called the line of cos a. Since tan a = AT/OA = AT/1 = AT, so AT is called the line of tan a. (Note: OA = OP = r = 1 ).

Example 2

The terminal side of angle a is pi/4. Find the value of x and y, sin a , cos a , and tan a.

Solution

In right triangle POM, since angle a = pi/4 = 45o, so its both legs have the same length. That is, PM = OM. Since r = 1, so PM = OM = Sqrt (2)/2. PM is called the line of sin a. OM is called the line of cos a. AT is called the line of tan a. (Note: OA = OP = r = 1 ).

Example 3

The terminal side of angle a is pi/3. Find the value of x and y, sin a , cos a , and tan a.

Solution

In right triangle POM, since angle a = pi/3 = 60o, so angle OPM = 30o, then the length of OM is one-half of the hypotenuse, that is, OM = r/2 , since r = 1 (given), so OM = 1/2. MP is called the line of sin a. OM is called the line of cos a. AT is called the line of tan a. (Note: OA = OP = r = 1 ).

The Value of Special Angles of Sine, Cosine, and Tangent

The following are the values of the special angles of sine, cosine, and tangent.

Trigonometric Functions of Any Angles

The following are the formula of trigonometric functions of any angles.

k is a integer, k = ... -5,-4,-3,-2,-1,0,1,2,3,4,5 ...

Example 4

Find the value of sin 960o.

Solution

Formula used:

sin(a + k × 360o) = sin(a)

sin(180o + a) = - sin(a)

The terminal side of the angle a is in quadrant III, so sin(a) is < 0.

Example 5

Find the value of cos ( - 600o).

Solution

cos(-600o)
= cos(-2 × 360o + 120o)
= cos(120o)
= cos(180o - 60o)
= - cos(60o)
= -1/2

Formula used:

cos(a + k × 360o) = cos a

cos(180o - a) = - cos a

The terminal side of the angle a is in quadrant II, so cos(a) is < 0.

Example 6

Find the value of tan 945o + sin 510o.

Solution
tan945o + sin510o
= tan(2 × 360o + 225o) + sin(1 × 360o + 150o)
= tan225o + sin150o
= tan(180o + 45) + sin(180o - 30o)
= tan45o + sin30o
= 1 + 1/2
= 3/2

Formula used:

tan(a + k × 360o) = tan a

In quadrant III, tan(a) > 0

In quadrant II, sin(a) > 0