values of the trig functions reference angles and inverse functions (5.4)

Values of the Trig Functions

Reference angles and inverse functions (5.4)

POD—introduction

Find the values of x for sin(x) = ½: (p. 406, #51)

In one positive rotation

In one negative rotation

For all rotations (general solution)

We’ll do this using the unit circle and the graphs.

POD—introduction

Find the values of x for sin(x) = ½:

In one positive rotation: π/6 and 5π/6

In one negative rotation: -11π/6 and -7π6(These are coterminal with the first answers.)

For all rotations (general solution): π/6 ± 2πn and 5π/6 ± 2πn (These are all the coterminal angles.)

POD—introduction

Now, find the values of x for sin(x) > ½ in one positive rotation:

POD—introduction

Now, find the values of x for sin(x) > ½ in one positive rotation:

between π/6 and 5π/6

Finally, find the values of x for sin(x) > ½ in one negative rotation.

POD—introduction

Now, find the values of x for sin(x) > ½ in one positive rotation:

Between π/6 and 5π/6

Finally, find the values of x for sin(x) > ½ in one negative rotation.

Between -7π6 and -11π/6 .

Reference angles

We often use the angles of the first quadrant of the unit circle as reference angles (θR) to determine exact trig values in all four quadrants.

Today we use reference angles in a slightly different context.

Trig inverses

When we find the angle of a given trig function, we are using inverse trig functions.

Makes sense, no? We work in the opposite direction to what we’ve been doing.

Trig inverses– notation

We can use two types of notation:

1. sin-1, cos-1, tan-1

(Here, the -1 is inverse function notation, not reciprocal notation.)

2. arcsin, arccos, arctan

These all read as “the angle which sine/ cosine/ tangent is (trig value here).”

Trig inverses– general idea

When we solve for angles, and use the inverse trig functions, we’ll get an angle answer.

We use the angle from the inverse trig function as a sort of reference angle.

Try it: find sin-1 .5

Trig inverses– graphs

Graph y = sin x in radian mode on your calculators.

If you were to anticipate the graph of its inverse, what would that look like?

Would that inverse be a function?

(What does that say about the graph of y = sin x?)

What could we do to make the inverse a function?

Trig inverses– graphs

We have to limit what we use of the original y = sin x.

We need to make sure we cover a domain that gives us a full range of sine from -1 to 1.

We should include the origin as well.

What do you suggest?

Trig inverses– graphs

Let’s limit it to –π/2 to π/2.

Compare the graphs.

Here we have y=sin x and y=sin-1x.

What are the endpoints?

Odd/ even/ neither?

Trig inverses– graphs

Now graph y = cos x and anticipate the inverse graph.

Endpoints?

Even/ odd/ neither?

You can see the inverse graphs for these graphs and tangent as well on the handout.

What are the “houses?”

Significance

So, while an infinite number of angles have a sine of .5,

sin-1 .5 = π/6 only.

An infinite number of angles have a cosine of .5, but

cos-1 .5 = π/3 only.

Significance

When we solve for angles, and use the inverse trig functions, we’ll get one answer, and may have to expand that answer to cover the full interval specified in the problem.

We use the angle from the inverse trig function as a sort of reference angle.

Try it.

Using calculators, find cos-1(-.3842) in a full rotation, 0 ≤ θ ≤ 2π.

In which quadrants will we find them?

Try it.

Using calculators, find cos-1(-.3842) in a full rotation.

We get θ = 1.9651. What quadrant is this in?

What angles would we have in a full rotation, 0 ≤ θ ≤ 2π? (Sometimes drawing a diagram is helpful.)

What angles would we have in general?

Try it.

Using calculators, find cos-1(-.3842). This will mean finding angles in the second and third quadrants, since cosine is negative. θ = 1.9651 is in the second quadrant.2π – θ = 4.3181 is the angle in the third quadrant.

In general, we’ll have these two angles and every angle coterminal to them:

1.9651 ± 2πn and 4.3181 ± 2πn

Try it again

If tan θ = -.4623 and 0 ≤ θ ≤ 360°, find θ to the nearest .1°.

Try it again

If tan θ = -.4623 and 0 ≤ θ ≤ 360°, find θ to the nearest .1°.

The calculators give θ = -24.8°. (Why?) This is out of given range, so we need to work the numbers. What do you suggest?

Try it again

If tan θ = -.4623 and 0 ≤ θ ≤ 360°, find θ to the nearest .1°.

Add 180° to -24.8°. What do you get?

Do we need to add it again?

Try it again

If tan θ = -.4623 and 0 ≤ θ ≤ 360°, find θ to the nearest .1°.

Final answer: θ = 155.2° and 335.2°.

How could you draft a general solution?

Try it again

If tan θ = -.4623 and 0 ≤ θ ≤ 360°, find θ to the nearest .1°.

Final answer: θ = 155.2° and 335.2°.

θ = 155.2° ±180°n θ = 335.2° ±180°n

Or in one elegant step: θ = 155.2° ±180°n