The corresponding acute angle

Let θ be an angle that terminates in any quadrant.  Then by the corresponding acute angle, we mean the shortest angular distance to the x-axis.

In each quadrant, φ is the corresponding acute angle.  It is the shortest angular distance to the x-axis.

Examples.   

If θ = 120° (Second quadrant), then φ = 60°.

If θ = 190° (Third quadrant), then φ = 10°.

If θ = 340° (Fourth quadrant), then φ = 20°.

Problem 1.   Name the corresponding acute angle.

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a)  110°   70°        b)  225°   45°        c)  −30°   30°

How, then, do we evaluate a function of any angle?  According to the following theorem:

A function of any angle will equal plus or minus that same function
of the corresponding acute angle.
The sign will depend on the quadrant.

Problem 2.   Draw a figure that illustrates the following.

The sine of an obtuse angle is equal to the sine of its supplement.

The cosine of an obtuse angle is equal to the negative of the cosine
of its supplement.

The supplement, 180° − θ, is the corresponding acute angle!

Problem 3.   Evaluate each of the following.  No tables.

a)  sin 150°  

The figure shows:  If θ terminates in the first quadrant at (a, b), then −θ will terminate at (a, −b).  Therefore,

But

Problem 5.   Use the previous theorem to evaluate the following.  No tables.

We can specify the position of a point P by giving its distance r from the origin and the angle θ it makes with the x-axis.  Those are called its polar coördinates.

But the polar coördinates are easily related to the rectangular coördinates (x, y).  Since

then

And since

Example 4.   A radius of 8 cm sweeps out an angle of 30° in standard position. What are the rectangular coordinates (x, y) of the endpoint of the radius?

Answer.		x = r cos θ  =  8 cos 30°  =  8·	2	=  4

A function of any angle will equal plus or minus that same function
of the corresponding acute angle.
The sign will depend on the quadrant.

First, if θ is a second quadrant angle, then r will terminate at a point
(−a, b).  The corresponding acute angle is φ, which is also shown in its first quadrant position.  In the second quadrant,

while in the first quadrant,

Therefore the sine of θ is equal to the sine of the corresponding acute angle.

Similarly,

And so on, for the remaining functions, so that in every case, a function of θ is plus or minus that same function of φ.

Next, if θ is a third quadrant angle, so that r terminates at (−a, −b), then

And so on, so that again, each function of θ is plus or minus that same function of φ.

Finally, if θ is a fourth quadrant angle, so that r terminates at (a, −b), then

Therefore again, each function of θ is plus or minus that same function of the corresponding acute angle φ;  which is what we wanted to prove.