Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Review: Expressing one trig functions in terms of a given trig function
Review example: Assume θ is an acute angle with cos θ = x.
Find the other five trig functions of θ.
Solution: We are given cos θ = x. To find out
other trig functions of θ, draw the acute triangle
A
at the right, with cos θ = H
= x1 = x as required.
Solve√A2 + O2 = H√2 for
O = H 2 − A2 = 1 − x 2 .
Now we can find the other trig functions of θ.
√
√
2
O
• sin θ = H
= 1−x
= 1 − x2 .
1
• tan θ =
• cot θ =
• sec θ =
• csc θ =
O
A
A
O
H
A
H
O
Hypotenuse = 1
p
1 − x2 = Opposite
θ
Adjacent = x
√
=
=
=
=
1−x2
.
x
1
√ x
tan θ = 1−x2 .
1
1
cos θ = x .
√ 1
.
1−x2
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Double-angle formulas
Recall the sum formulas:
• cos(u + v) = cos(u) cos(v) − sin(u) sin(v)
• sin(u + v) = sin(u) cos(v) + cos(u) sin(v)
tan u+tan v
• tan(u + v) = 1−tan
u tan v
In each formula, substitute u for v.
Double-angle formulas.
Let u be any angle , expressed in radians or degrees. Then:
• cos(2u) = cos2 u − sin2 u • sin(2u) = 2 sin u cos u • tan(2u) =
2 tan u
1−tan2 u
Solving cos2 u + sin2 u = 1 for cos2 θ = 1 − sin2 θ yields an alternate double angle
formula for cosine: cos(2u) = cos2 u − sin2 u = 1 − sin2 θ − sin2 θ = 1 − 2 sin2 θ.
Solving cos2 u + sin2 u = 1 for sin2 θ = 1 − cos2 θ yields cos(2u) = 2 cos2 u − 1.
More double-angle formulas for cosine
Let u be any angle , expressed in radians or degrees. Then:
• cos(2u) = 2 cos2 u − 1
• cos(2u) = 1 − 2 sin2 u
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Using the double-angle formulas
Example 1: Suppose cos θ = − 23 and θ is in Quadrant 3. Find cos(2θ) and sin(2θ).
Solution:
• To find cos(2θ), use the double-angle formula and the identity sin2 θ = 1 − cos2 θ.
Then cos(2θ) = cos2 θ − sin2 θ = cos2 θ − (1 − cos2 θ) = 2 cos2 θ − 1 = 2(− 23 )2 − 1 =
2( 49 ) − 1 =
8
9
−
9
9
= − 19 .
• To find sin(2θ), use the double-angle formula sin(2θ) = 2 sin θ cos θ. We are given
cos θ = − 23 but need to figure out sin θ.
√
√
For this you need to remember that if A2 = 7, then A = ± 7, NOT 7.
Since sin2 θ = 1 − cos2 θ, we
q have
q
q
√
√
2
sin θ = ± 1 − cos θ = ± 1 − (− 23 )2 = ± 1 − 49 = ± 59 ± 35 .
To decide between plus and minus, use
√ the fact that θ is a Quadrant 3 angle, and so its
sine is negative. Therefore sin θ = − 35 . Now we can apply the double-angle formula:
√
sin(2θ) = 2 sin θ cos θ = 2(−
5
2
3 )(− 3 )
√
= −495 .
If these answers are correct, then cos2 2θ + sin2 2θ should be 1. Please check that this is so.
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Half-angle formulas
We now combine the identities cos 2θ = cos2 θ − sin2 θ and cos2 θ + sin2 θ = 1
Adding them yields 2 cos2 θ = 1 + cos(2θ). Dividing by 2 gives cos2 θ = 1+cos(2θ)
.
2
1−cos(2θ)
2
2
2
2
2
Subtracting cos 2θ = cos θ − sin θ from cos θ + sin θ = 1 yields sin θ =
2
Formulas for lowering powers
Let θ be any angle , expressed in radians or degrees. Then:
2θ
2θ
• sin2 θ = 1−cos
• tan2 θ =
• cos2 θ = 1+cos
2
2
Substituting
u
2
for θ and solving for the three trig functions of
u
2
1−cos 2θ
1+cos 2θ
yields
Half-angle formulas
Let u be anyqangle , expressed in radians
q or degrees. Then:
q
u
u
u
1+cos u
1−cos u
1−cos u
•
sin
=
±
•
tan
=
±
• cos 2 = ±
2
2
2
2
1+cos u
The choice of sign depends on which quadrant angle u2 is in.
You need to know and apply these six formulas. Rather than memorizing the formulas
themselves, memorize how to derive them, as shown above.
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Using the half-angle formulas
Example 2: Find the exact value of sin π8 .
45
◦
Solution: Since π radians = 180◦ , the angle is 180
8 = 2 = 22.5 . The .5 should alert
you to the fact that you need to use a half-angle formula for sin( u2 ). In other words,
set u2 = 22.5◦ and solve for u to obtain u = 45◦ .
r
q
q
1− √1
u
1−cos 45◦
2
=
±
=
±
Then sin 22.5◦ = sin u2 = ± 1−cos
2
2
2 . To figure out the
sign, note that 22.5◦ris in Quadrant 1, where sine is positive. Therefore
1− √1
2
sin π8 = sin 22.5◦ =
You can make the answer look nicer by multiplying top
2
and bottom
the radical by 2.
r under
r of the fraction
q √
1
1
p
√
√
√
2
1−
1−
2
2− 2
π
1
2
Answer:
=
=
=
sin
=
2
−
2.
2
2·2
4
8
2
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Example 3: Find the exact value of cos 112.5◦ .
Solution: The .5 should alert you to the fact that you need to use a half-angle formula
for cos( u2 ). In other words, set u2 = 112.5◦ and solve for u to obtain u = 225◦ .
q
q
u
1+cos 225◦
Then cos 112.5◦ = cos u2 = ± 1+cos
=
±
2
2
To proceed further, we need to pause and figure out cos 225◦ . First, u = 225◦ is a Quadrant 3
angle. Draw a third quadrant picture to see that the reference angle of u = 225◦ is
225◦ − 180◦ = 45◦ = u. Now recall the Reference Angle Principle: cos u = ± cos u. The sign is
negative because cosine is negative in Quadrant 3. Therefore, setting u = 225◦ , we obtain
cos 225◦ = − cos(45◦ ) = − √12 .
r
q
1− √1
◦
1+cos
225
2
Now we go back and calculate cos 112.5◦ = ±
=±
2
2 .
◦
The sign is negative since
r 112.5 is a Quadrant 2 angle, where cosine is negative.
Thus cos 112.5◦ = −
1− √1
2
. Multiplying top and bottom of the fraction under the
q √
p
√
radical by 2 gives the neater form − 2−4 2 = cos 112.5◦ = − 12 2 − 2 .
2
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Example 4: Find cos u2 if sin u = − 35 and u is in Quadrant 3.
Solution:
q
u
.
Begin with the half-angle formula cos u2 = ± 1+cos
2
First figure out cos u. Sinceq
cos2 u + sin2 u =q
1, solve for q
p
2
9
4
cos u = ± 1 − sin2 u = ± 1 − − 35 = ± 1 − 25
= ± 16
25 = ± 5 .
Since u is in Quadrant 3, where cosine is negative, cos u = − 45 .
q 4
q
q
1− 5
u
1+cos u
1
The half-angle formula tells us cos 2 = ±
=
±
=
±
2
2
10 . To figure out
u
the sign, we need to be careful and find out what quadrant 2 is in. Since u is in
Quadrant 3, 180◦ < u < 270◦ . Divide by 2 to get 90◦ < u2 < 135◦ .
q
1
.
Therefore u2 is in Quadrant 2, where cosine is negative, and so cos u2 = − 10
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Example 5: Find cos(2 arccos(x)).
Solution: Let θ = arccos(x). Then cos θ = x. Then
cos 2θ = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 2x2 − 1 .
Example 6. Find tan(2 arcsin x).
Solution: Let θ = arcsin x. Then sin θ = x. We want to calculate tan 2θ =
We only know sin θ = x. To find out tan θ or other
trig functions of θ, draw the acute triangle at the
O
= x1 = x as required.
right, with sin θ = H
√
√
2
2
2
Solve A + O = H for A = H 2 − O2 = 1 − x2 .
√ x
Then tan θ = O
.
A =
1−x2
Hypotenuse = 1
2 tan θ
.
1−tan2 θ
x = Opposite
θ
Adjacent =
p
1 − x2
2 tan θ
Now we are ready to find tan 2θ = 1−tan
2 θ.
x
x2
1−x2
x2
1−2x2
• Find the denominator 1 − tan2 θ = 1 − ( √1−x
)2 = 1 − 1−x
2 = 1−x2 − 1−x2 = 1−x2 .
2
2x
• Find the numerator 2 tan θ = √1−x
.
2
√
√ 2x
√
2 tan θ
2x·( 1−x2 )2
1−x2
2x 1−x2
1−x2
√ 2x
√
=
=
Then tan 2θ =
=
·
=
.
1−2x2
1−2x2
1−x2 1−2x2
( 1−x2 )·(1−2x2 )
1 − tan2 θ
2
1−x
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas
Review
Double-angle formulas
Half-angle formulas
Inverse function examples
Quiz Review
Quiz Review
Review example: Assume θ is an acute angle. If cos θ = x, find the other five trig
functions of θ.
Example 1: Suppose cos θ = − 32 and θ is in Quadrant 3. Find cos(2θ) and sin(2θ).
Example 2: Find the exact value of sin π8 .
Example 3: Find the exact value of cos 112.5◦ .
Example 4: Find cos u2 if sin u = − 35 and u is in Quadrant 3.
Example 5: Find cos(2 arccos(x)).
Example 6. Find tan(2 arcsin x).
Stanley Ocken
M19500 Precalculus Chapter 7.3 Double and half-angle formulas