4.9 Solving Trig Equations Using the Pythagorean Identities

4.9 Solving Trig Equations Using the Pythagorean Identities

4.9.1 The Pythagorean Identities From the Pythagorean theorem we found the equation for the unit circle:

x2 + y2 = 1.

From that equation and from our definition of cos as the x-value and sin as the y-value of points on

the circle, we discovered the identity

cos2 + sin2 = 1.

(15)

If we divide both sides of equation 15 above by cos2 we can rewrite that identity as

1 + tan2 = sec2 .

(16)

If we divide both sides of equation 15 above by sin2 we can rewrite that identity as

cot2 + 1 = csc2 .

(17)

This triplet of identities are called the Pythagorean identities. We will often use them to change even powers of one trig function into an expression involving even powers of another.

Here are some examples.

1. Solve cos2 x + 5 sin x - 7 = 0. Solution. We replace cos2 x by 1 - sin2 x and so consider the equation

1 - sin2 x + 5 sin x - 7 = 0.

This simplifies to

- sin2 x + 5 sin x - 6 = 0.

Multiply both sides by -1 to get the equation

sin2 x - 5 sin x + 6 = 0.

Factor the lefthand side:

(sin x - 2)(sin x - 3) = 0

and note that this requires sin x = 2 or sin x = 3, both of which are impossible. So there is NO solution to this equation

2. Solve the equation sec2 + tan - 1 = 0

Solution. Suppose sec2 + tan - 1 = 0. Using a Pythagorean identity we replace sec2 by tan2 + 1 and write this as

(tan2 + 1) + tan - 1 = 0.

Simplify, so that

tan2 + tan = 0.

Factor out tan to get

tan (tan + 1) = 0.

So either

tan = 0

or tan = -1.

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3 In the first case, = 0 + k and in the second case, = + k. So our answer is

4 3 k, or + k 4 where k is an integer.

3. Solve the equation tan2 + 3 sec + 3 = 0. Solution. We replace tan2 by sec2 - 1 and solve for sec :

(sec2 - 1) + 3 sec + 3 = 0

Therefore either or

= sec2 + 3 sec + 2 = 0 = (sec + 2)(sec + 1) = 0

sec + 2 = 0 sec + 1 = 0.

Case 1.

1 sec + 2 = 0 = sec = -2 = cos = -

2 = ( = 2/3 + 2k) or ( = 4/3 + 2k.)

Case 2.

sec + 1 = 0 = sec = -1 = cos = -1 = = + 2k

In our final solution we bring all these together:

= 2/3 + 2k or = 4/3 + 2k or = + 2k

(where k Z.)

4. Solve 3 cos x = sin x + 1.

Solution. Since cosine and sine are both in this equation, it would be nice if they were squared. We can use that idea by first squaring both sides:

( 3 cos x)2 = (sin x + 1)2

and so

3 cos2 x = sin2 x + 2 sin x + 1.

Replace cos2 x by 1 - sin2 x to get the equation

3(1 - sin2 x) = sin2 x + 2 sin x + 1.

so (after moving everything to the righthand side) we have 0 = 4 sin2 x + 2 sin x - 2.

Divide both sides by 2

0 = 2 sin2 x + sin x - 1

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and factor 2 sin2 x + sin x - 1 = (2 sin x - 1)(sin x + 1). Thus we have

0 = (2 sin x - 1)(sin x + 1)

and

so

either

sin x

=

1 2

or

sin x

=

-1.

In

the

first

case

x = /6 + 2k or x = 5/6 + 2k

and in the second case

x = -/2 + 2k.

Now when we squared both sides we may have introduced additional elements into our solution set (for example, maybe we now have elements where sin x + 1 = - 3 cos x) and so we should check

our answers. (It is always good practice to check our answers!)

If

x

=/6

then

sin x + 1

=

3 2

which

is

the

same

as

3 cos(/6).

But

if

x

=

5/6

then

sin x + 1

=

3 2

while

3 cos(5/6)

=

-

3 2

.

So

x

=

5/6

is

not a

solution

to

the

original

equation!

Finally, if x = -/2 then sin x + 1 = 0 = 3 cos x, so x = -/2 is a solution to the original

equation. So our final solution set is

{/6 + 2k : k Z} {-/2 + 2k : k Z} .

4.9.2 Showing something is not an identity

To show that an equation is not an identity it suffices to find just one value of for which the equation fails!

For example, is the equation

sin = cos

an identity? No ? try = 0 and see that the lefthand side is not the same as the righthand side.

Here is another example: Is

cos = 1 - sin2

an identity? Here we might want to try an angle like = to see that this equation can fail (since the lefthand side is negative while the righthand side is positive.)

Homework. As class homework, finish off Worksheet 4.9, Trig Identities and Equations available through

the class webpage.

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