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 θ as the x-value and sin θ as the y-value of points onthe circle, we discovered the identity

cos2 θ + sin2 θ = 1. (15)

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

1 + tan2 θ = sec2 θ. (16)

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

cot2 θ + 1 = csc2 θ. (17)

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

Here are some examples.

1. Solve cos2 x+ 5 sinx− 7 = 0.

Solution. We replace cos2 x by 1− sin2 x and so consider the equation

1− sin2 x+ 5 sinx− 7 = 0.

This simplifies to− sin2 x+ 5 sinx− 6 = 0.

Multiply both sides by −1 to get the equation

sin2 x− 5 sinx+ 6 = 0.

Factor the lefthand side:(sinx− 2)(sinx− 3) = 0

and note that this requires sinx = 2 or sinx = 3, both of which are impossible. So there is

NO solution to this equation

2. Solve the equation sec2 θ + tan θ − 1 = 0

Solution. Suppose sec2 θ + tan θ − 1 = 0. Using a Pythagorean identity we replace sec2 θ bytan2 θ + 1 and write this as

(tan2 θ + 1) + tan θ − 1 = 0.

Simplify, so thattan2 θ + tan θ = 0.

Factor out tan θ to gettan θ(tan θ + 1) = 0.

So eithertan θ = 0

ortan θ = −1.

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In the first case, θ = 0 + πk and in the second case, θ =3π

4+ πk. So our answer is

πk, or3π

4+ πk

where k is an integer.

3. Solve the equation tan2 θ + 3 sec θ + 3 = 0.

Solution. We replace tan2 θ by sec2 θ − 1 and solve for sec θ :

(sec2 θ − 1) + 3 sec θ + 3 = 0

=⇒ sec2 θ + 3 sec θ + 2 = 0

=⇒ (sec θ + 2)(sec θ + 1) = 0

Therefore eithersec θ + 2 = 0

orsec θ + 1 = 0.

Case 1.

sec θ + 2 = 0 =⇒ sec θ = −2 =⇒ cos θ = −1

2

=⇒ (θ = 2π/3 + 2kπ) or (θ = 4π/3 + 2kπ.)

Case 2.sec θ + 1 = 0 =⇒ sec θ = −1 =⇒ cos θ = −1 =⇒ θ = π + 2kπ

In our final solution we bring all these together:

θ = 2π/3 + 2kπ or θ = 4π/3 + 2kπ or θ = π + 2kπ

(where k ∈ Z.)

4. Solve√

3 cosx = sinx+ 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 cosx)2 = (sinx+ 1)2

and so3 cos2 x = sin2 x+ 2 sinx+ 1.

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

3(1− sin2 x) = sin2 x+ 2 sinx+ 1.

so (after moving everything to the righthand side) we have

0 = 4 sin2 x+ 2 sinx− 2.

Divide both sides by 20 = 2 sin2 x+ sinx− 1

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and factor 2 sin2 x+ sinx− 1 = (2 sinx− 1)(sinx+ 1). Thus we have

0 = (2 sinx− 1)(sinx+ 1)

and so either sinx = 12 or sinx = −1. In the first case

x = π/6 + 2πk or x = 5π/6 + 2πk

and in the second casex = −π/2 + 2πk.

Now when we squared both sides we may have introduced additional elements into our solution set(for example, maybe we now have elements where sinx + 1 = −

√3 cosx) and so we should check

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

If x = π/6 then sinx+ 1 = 32 which is the same as

√3 cos(π/6). But if x = 5π/6 then sinx+ 1 = 3

2

while√

3 cos(5π/6) = − 32 . So x = 5π/6 is not a solution to the original equation!

Finally, if x = −π/2 then sinx + 1 = 0 =√

3 cosx, so x = −π/2 is a solution to the originalequation. So our final solution set is

{π/6 + 2πk : k ∈ Z} ∪ {−π/2 + 2πk : 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 θ for which the equationfails!

For example, is the equationsin θ = cos θ

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

Here is another example: Is

cos θ =√

1− sin2 θ

an identity? Here we might want to try an angle like θ = π to see that this equation can fail (since thelefthand 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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