### Section 9.1

```9.1
IDENTITIES, EXPRESSIONS,
AND EQUATIONS
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Deriving New Identities
Example 1
The Pythagorean identity, cos2 θ + sin2 θ = 1, can be rewritten in
terms of other trigonometric functions. Provided cos θ ≠ 0,
dividing through by cos2 θ gives
cos2  sin 2 
1


cos2  cos2  cos2 
 sin    1 
1 
 

 cos   cos 
2
2
The identity we derived is another version of the Pythagorean
identity. It is usually written
1 + tan2 θ = sec2 θ.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Using Identities To Simplify Expressions
Example 3
Simplify the expression
(2 cos t + 3 sin t) (3 cos t + 2 sin t) − 13 sin t cos t.
Solution
To make the calculations easier, let r = cos t and s = sin t. Our
expression becomes
(2r + 3s)(3r + 2s) − 13rs = 6r2 + 4rs + 9rs + 6s2 − 13rs multiply out
= 6r2 + 6s2 + 13rs − 13rs
regroup
= 6 (r2 + s2)
simplify and factor
= 6 (cos2 t + sin2 t)
r = cos t, s = sin t
= 6.
because cos2 t + sin2 t = 1
We see that this complicated expression equals 6.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Using Identities to Evaluate Expressions
Example 3
Suppose that cos θ = 2/3 and 3π/2 ≤ θ ≤ 2π. Find sin θ and tan θ
Solution
Use the relationship cos2 θ + sin2 θ = 1 to find sin θ.
Substitute cos θ = 2/3:
(2/3)2 + sin2 θ = 1
4/9 + sin2 θ = 1
sin2 θ = 1 − 4/9 = 5/9
sin θ =  5 / 9   5 / 3
Because θ is in the fourth quadrant, sin θ is negative, so
sin θ = − 5 / 3. To find tan θ, use the relationship
tan θ = sin θ / cos θ = (− 5 / 3 )/(2/3) = − 5 / 2
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Using Identities to Solve Equations
Example 6
Solve 2 sin2 t = 3 − 3 cos t for 0 ≤ t ≤ π.
Solution
We can use the identity sin2 t = 1 − cos2 t to rewrite this equation
entirely in terms of cos t:
2 sin2 t = 2(1 − cos2 t) = 2− 2cos2 t = 3 − 3 cos t
2 cos2 t − 3 cos t + 1 = 0.
It can be helpful to abbreviate cos t = x, so x2 = (cos t)2 = cos2 t,
giving:
2x2 − 3x + 1 = 0
(2x − 1)(x − 1) = 0
so x = ½ or x = 1. Since x = cos t, this means either cos t = ½ with
solution t = π /3, or cos t = 1 with solution t = 0.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Double-Angle Formula for Sine
Thinking algebraically, we would like to find a formula for sin 2θ in
terms of sin θ and cos θ. We derive our formula by using the figure.
The lengths of OA and OC are 1; the length of AC is 2sin θ. Writing
for the angle at A and applying the Law of Sines to triangle OAC
gives
(sin 2θ) / (2 sin θ) = sin α / 1
In triangle OAB, the length of side OB
A
is cos θ, and the hypotenuse is 1, so
1
α
sin θ
sin α = Opposite / Hypotenuse = cos θ / 1 = cos θ
θ
θ cos θ B
O
Thus, substituting cos θ for sin α, we have
sin θ
1
(sin 2θ)/(2 sin θ) = cos θ
C
sin 2θ = 2 sin θ cos θ
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Using Identities to Solve Equations
Example 7
Find all solutions to the equation sin 2t = 2 sin t on the interval
0 ≤ t ≤ 2π.
Solution
Using the double-angle formula sin 2t = 2 sin t cos t, we have
2 sin t cos t = 2 sin t
2 sin t cos t − 2 sin t = 0
2 sin t (cos t − 1) = 0
factoring out 2 sin t.
Thus,
2 sin t = 0 or cos t = 1.
Now, 2 sin t = 0 for t = 0, π, and 2 π, and cos t = 1 for t = 0 and 2 π.
Thus there are three solutions to the original equation on the
interval 0 ≤ t ≤ 2 π : t = 0, π, and 2 π.
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Using Identities to Solve Equations
Example 7 (continued)
Find all solutions to the equation sin 2t = 2 sin t on the interval
0 ≤ t ≤ 2π.
Solution
The figure illustrates the solutions (t = 0, π, and 2 π) graphically as
the points where the graphs of y = sin 2t and y = 2 sin t intersect.
y = 2 sin t
y = sin 2t
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
Double-Angle Formulas
for Cosine and Tangent
The double-angle formula for the
cosine can be written in three forms:
cos 2θ = 1 − 2 sin2θ
cos 2θ = 2 cos2θ − 1
cos 2θ = cos2θ − sin2θ
tan 2θ = (2 tan θ) / (1 − tan2 θ)
Functions Modeling Change:
A Preparation for Calculus,
4th Edition, 2011, Connally
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