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 + y 2 = 1.
From that equation and from our definition of cos θ as the x-value and sin θ as the y-value of points on
the 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 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 θ + tan θ − 1 = 0
Solution. Suppose sec2 θ + tan θ − 1 = 0. Using a Pythagorean identity we replace sec2 θ by
tan2 θ + 1 and write this as
(tan2 θ + 1) + tan θ − 1 = 0.
Simplify, so that
tan2 θ + tan θ = 0.
Factor out tan θ to get
tan θ(tan θ + 1) = 0.
So either
tan θ = 0
or
tan θ = −1.
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In the first case, θ = 0 + πk and in the second case, θ =
πk, or
3π
+ πk. So our answer is
4
3π
+ πk
4
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 either
sec θ + 2 = 0
or
sec θ + 1 = 0.
Case 1.
1
2
=⇒ (θ = 2π/3 + 2kπ) or (θ = 4π/3 + 2kπ.)
sec θ + 2 = 0 =⇒ sec θ = −2 =⇒ cos θ = −
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 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 = π/6 + 2πk or x = 5π/6 + 2πk
and in the second case
x = −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 sin x + 1 = − 3 cos x) and so we should check
our answers. (It is always good practice to check our answers!)
√
If x = π/6 then sin x + 1 = 23 which is the same as 3 cos(π/6). But if x = 5π/6 then sin x + 1 = 32
√
while 3 cos(5π/6) = − 23 . So x = 5π/6 is not a solution to the original equation!
√
Finally, if x = −π/2 then sin x + 1 = 0 = 3 cos x, so x = −π/2 is a solution to the original
equation. 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 equation
fails!
For example, is the equation
sin 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 =
p
1 − sin2 θ
an identity? Here we might want to try an angle like θ = π 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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