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Complementary angles are positive acute angles that add up to 90 . e.g. 20 and 70 are complementary angles.

Sine and cosine are known as co-functions . Two functions are called co-functions if f ( A ) = g ( B ) whenever A + B = 90 (i.e. A and B are complementary angles). The other trig co-functions are secant and cosecant, and tangent and cotangent.

The function value of an angle is equal to the co-function of its complement (the co-co rule).

Thus for sine and cosine we have

sin ( 90 - θ ) = cos θ cos ( 90 - θ ) = sin θ

Write each of the following in terms of 40 using sin ( 90 - θ ) = cos θ and cos ( 90 - θ ) = sin θ .

  1. cos 50
  2. sin 320
  3. cos 230
    1. cos 50 = cos ( 90 - 40 ) = sin 40
    2. sin 320 = sin ( 360 - 40 ) = - sin 40
    3. cos 230 = cos ( 180 + 50 ) = - cos 50 = - cos ( 90 - 40 ) = - sin 40

Function values of ( θ - 90 )

sin ( θ - 90 ) = - cos θ and cos ( θ - 90 ) = sin θ .

These results may be proved as follows:

sin ( θ - 90 ) = sin [ - ( 90 - θ ) ] = - sin ( 90 - θ ) = - cos θ

and likewise for cos ( θ - 90 ) = sin θ

Summary

The following summary may be made

second quadrant ( 180 - θ ) or ( 90 + θ ) first quadrant ( θ ) or ( 90 - θ )
sin ( 180 - θ ) = + sin θ all trig functions are positive
cos ( 180 - θ ) = - cos θ sin ( 360 + θ ) = sin θ
tan ( 180 - θ ) = - tan θ cos ( 360 + θ ) = cos θ
sin ( 90 + θ ) = + cos θ tan ( 360 + θ ) = tan θ
cos ( 90 + θ ) = - sin θ sin ( 90 - θ ) = sin θ
cos ( 90 - θ ) = cos θ
third quadrant ( 180 + θ ) fourth quadrant ( 360 - θ )
sin ( 180 + θ ) = - sin θ sin ( 360 - θ ) = - sin θ
cos ( 180 + θ ) = - cos θ cos ( 360 - θ ) = + cos θ
tan ( 180 + θ ) = + tan θ tan ( 360 - θ ) = - tan θ
  1. These reduction formulae hold for any angle θ . For convenience, we usually work with θ as if it is acute, i.e. 0 < θ < 90 .
  2. When determining function values of 180 ± θ , 360 ± θ and - θ the functions never change.
  3. When determining function values of 90 ± θ and θ - 90 the functions changes to its co-function (co-co rule).

Function values of ( 270 ± θ )

Angles in the third and fourth quadrants may be written as 270 ± θ with θ an acute angle. Similar rules to the above apply. We get

third quadrant ( 270 - θ ) fourth quadrant ( 270 + θ )
sin ( 270 - θ ) = - cos θ sin ( 270 + θ ) = - cos θ
cos ( 270 - θ ) = - sin θ cos ( 270 + θ ) = + sin θ

Solving trigonometric equations

In Grade 10 and 11 we focussed on the solution of algebraic equations and excluded equations that dealt with trigonometric functions (i.e. sin and cos ). In this section, the solution of trigonometric equations will be discussed.

The methods described in previous Grades also apply here. In most cases, trigonometric identities will be used to simplify equations, before finding the final solution. The final solution can be found either graphically or using inverse trigonometric functions.

Graphical solution

As an example, to introduce the methods of solving trigonometric equations, consider

sin θ = 0 , 5

As explained in previous Grades,the solution of Equation [link] is obtained by examining the intersecting points of the graphs of:

y = sin θ y = 0 , 5

Both graphs, for - 720 < θ < 720 , are shown in [link] and the intersection points of the graphs are shown by the dots.

Plot of y = sin θ and y = 0 , 5 showing the points of intersection, hence the solutions to the equation sin θ = 0 , 5 .

In the domain for θ of - 720 < θ < 720 , there are 8 possible solutions for the equation sin θ = 0 , 5 . These are θ = [ - 690 , - 570 , - 330 , - 210 , 30 , 150 , 390 , 510 ]

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Source:  OpenStax, Math 1508 (lecture) readings in precalculus. OpenStax CNX. Aug 24, 2011 Download for free at http://cnx.org/content/col11354/1.1
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