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!
Elementary Functions
For example, is the equation
Part 4, Trigonometry
Lecture 4.9a, Showing an Equation is Not an Identity
sin θ = cos θ
an identity? No – try θ = 0 and see that the lefthand side is not the same
as the righthand side.
Dr. Ken W. Smith
Sam Houston State University
0 6= 1
2013
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Elementary Functions
2013
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Showing something is not an identity
Smith (SHSU)
Elementary Functions
2013
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Showing something is not an identity
Is 2 cos x = cos 2x?
Let’s try x = 0.
Is
cos θ =
p
1 − sin2 θ
2 cos 0 = 2(1) = 2
an identity?
cos(2 · 0) = cos 0 = 1
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.)
2 6= 1!
Is 2 sin x = sin 2x?
−1 6= 1
If I try x = 0, I get a true statement. But this just means I was unlucky....
Try x = π/2. (Always pick something fairly simple to compute.)
2 6= 0
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Elementary Functions
2013
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Elementary Functions
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Equations which are not identities
Elementary Functions
Part 4, Trigonometry
Lecture 4.9b, The Pythagorean Identities
In the next presentation, we will look in depth at the Pythagorean
Identities.
Dr. Ken W. Smith
(End)
Sam Houston State University
2013
Smith (SHSU)
Elementary Functions
2013
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Smith (SHSU)
Elementary Functions
The Pythagorean Identities
The Pythagorean Identities
From the Pythagorean theorem we found the equation for the unit circle:
Solve cos2 x + 5 sin x − 7 = 0.
x2 + y 2 = 1.
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Solution. We replace cos2 x by 1 − sin2 x and so consider the equation
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.
cos2 θ
If we divide both sides of equation 1 above by
identity as
1 + tan2 θ = sec2 θ.
1 − sin2 x + 5 sin x − 7 = 0.
(1)
This simplifies to
− sin2 x + 5 sin x − 6 = 0.
we can rewrite that
Multiply both sides by −1 to get the equation
(2)
sin2 x − 5 sin x + 6 = 0.
If we divide both sides of equation 1 above by sin2 θ we can rewrite that
identity as
cot2 θ + 1 = csc2 θ.
(3)
Factor the lefthand side:
This triplet of identities are called the Pythagorean identities. We use
them to change even powers of one trig function into an expression
involving
even powers of another.
Smith (SHSU)
Elementary Functions
2013
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.
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(sin x − 2)(sin x − 3) = 0
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Elementary Functions
2013
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The Pythagorean Identities
The Pythagorean Identities
Solve the equation sec2 θ + tan θ − 1 = 0
Solve the equation tan2 θ + 3 sec θ + 3 = 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
Solution. We replace tan2 θ by sec2 θ − 1 and solve for sec θ :
2
(sec2 θ − 1) + 3 sec θ + 3 = 0 =⇒ sec2 θ + 3 sec θ + 2 = 0
(tan θ + 1) + tan θ − 1 = 0.
Simplify, so that
=⇒ (sec θ + 2)(sec θ + 1) = 0
tan2 θ + tan θ = 0.
Factor out tan θ to get
=⇒ sec θ + 2 = 0 or sec θ + 1 = 0.
tan θ(tan θ + 1) = 0.
Case 1. sec θ + 2 = 0 =⇒ sec θ = −2 =⇒ cos θ = − 12
So either
=⇒ (θ = 2π/3 + 2kπ) or (θ = 4π/3 + 2kπ.)
tan θ = 0
or
Case 2.
sec θ + 1 = 0 =⇒ sec θ = −1 =⇒ cos θ = −1 =⇒ θ = π + 2kπ
tan θ = −1.
In the first case, θ = 0 + πk and in the second case, θ =
answer is
πk
or 3π
+ πk
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Elementary
4Functions
3π
4
+ πk. So our
2013
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where k is an integer.
The Pythagorean Identities
Solve
√
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
Elementary Functions
2013
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√
We attacked the equation 3 cos x = sin x + 1 by squaring both sides
and using a Pythagorean identity to change the equation into
0 = (2 sin x − 1)(sin x + 1)
and so either sin x =
In the first case
1
2
or sin x = −1.
x = π/6 + 2πk or x = 5π/6 + 2πk
3 cos2 x = sin2 x + 2 sin x + 1.
Replace cos2 x by 1 − sin2 x to get the equation
and in the second case
x = −π/2 + 2πk.
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
and factor 2 sin2 x + sin x − 1 = (2 sin x − 1)(sin x + 1). Thus we have
Elementary
0 = (2 sin
x − Functions
1)(sin x + 1)
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The Pythagorean Identities
3 cos x = sin x + 1.
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θ = 2π/3 + 2kπ or θ = 4π/3 + 2kπ or θ = π + 2kπ
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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 = 32 which is the same as 3 cos(π/6). But if
√
x = 5π/6 then sin x + 1 = 32 while 3 cos(5π/6) = − 32 . So x = 5π/6 is
not a solution to the original equation!
our final
solution set is
The other
check out so
Smith solutions
(SHSU)
Elementary
Functions
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Using the Pythagorean Identity
In the next presentations we examine some more trig identities.
(End)
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