W13D2 Tan graph
Warm Up Create
π
1. Evaluate tan
2
2. Evaluate tan(π)
π
3. Evaluate tan
4
π
4. Evaluate tan 2 · (
4
Lesson 29 Graphing Tangent
x
tan(x)
0
0
π
4
π
2
3π
4
π
5π
4
3π
2
7π
4
2π
3
2.5
2
1.5
1
0.5
π π 3π π 5π 3π 7π 2π 9π 5π
4 2 4
4 2 4
4 2
−0.5
−1
−1.5
−2
−2.5
−3
π 3π 5π
Asymptotes at ,
,
...
2 2 2
Period π
Points on the Midline 0, π, 2π, 3π.....
EX 2:
Amp - none
Midline = 2
Period
π
3
Graph tan(3x) + 2
1
undef
−1
0
1
undef
1
0
π 1
π
first Asymptote = · =
2 3
6
π π
All Asymptotes:
+ n
6
3
Midline Points 0, π, 2π, 3π.....
4
3.5
3
2.5
2
1.5
1
0.5
−0.5
π
6
π
3
π
2
2π
3
−1
Figure 1: tan(3x)
EX 3:
You Try y = tan
x
4
−1
Midline = -1
Period 4π
Asymptotes at 2π, 6π, 10π...
Midline Points 0, π, 2π, 3π.....
3
2.5
2
1.5
1
0.5
−0.5
−1
−1.5
−2
−2.5
−3
π π 3π 2π 5π 3π 7π 4π 9π 5π11π6π13π7π15π
2
2
2
2
2
2
2
2
Figure 2: 4 tan(x)
π
y = −4 tan 2( x +
3
EX 4:
Midline = 0
π
6π
=
2
12
4π
π
h=− =−
3
12
π 1 π
3π 4π
Asymptotes = · − =
−
2 2
3
12
12
π
First Asymptote = −
12
π
6π
5π
Asymptotes = − +
n=
,
12
12
12
Period =
11π
,
12
17π
12
3
2.5
2
1.5
1
0.5
−0.5
−1
−1.5
−2
−2.5
−3
π
6
π
3
π
2
2π 5π
3 6
π
7π 4π 3π
6 3 2
Figure 3: 4 tan(x)
Lesson 30 Trig Identities
1
θ
2
y
x
2
x +y =1
1
sin θ
θ
cos θ
Substitute
(cos θ)2 + (sin θ)2 = 1
This is the first Pythagorean Identity
Divide all by cos2 θ
cos2 θ
1
sin2 θ
+
=
cos2 θ cos2 θ
cos2 θ
tan2 θ + 1 = sec2 θ
Divide by sin2 θ
1
sin2 θ cos2 θ
+
=
2
2
sin θ
sin θ
sin2 θ
1 + cot2 θ = csc2 θ
You might want to make a list of identities and tips for using them
Pythagorean Identities
(cos θ)2 + (sin θ)2 = 1
tan2 θ + 1 = sec2 θ
1 + cot2 θ = csc2 θ
Reciprocal Identities
EX 1:
If sin θ =
tan θ =
sin θ
cos θ
csc θ =
1
sin θ
sec θ =
1
cos θ
cot θ =
cos θ
sin θ
3
π
and ≤ θ ≤ π solve for the other 5 trig functions
7
2
7
3
θ
cos θ
3
(cos θ)2 + ( )2 = 1
7
3
(cos θ) = 1 − ( )2 =
7
2
√
√
2 10
40
=
7
7
In Q2 ccsine is negative
√
2 10
cos θ = −
7
3
√
sin θ
3
3 10
tan θ =
= − √7 = − √ = −
cos θ
20
2 10
2 10
7
7
csc θ =
√ 3
7 10
sec θ = −
20
√
2 10
cot θ = −
7
EX 2:
if tan θ =
25
34
+ 1 = sec2 θ =
9
9
√
3 34
cos θ = −
34
3
cot θ =
5
5
3π
and π ≤ θ ≤
solve for the other 5 trig functions
3
2
√
34
sec θ = −
3
Use sin2 θ + cos2 θ = 1 or tan θ =
sin θ
cos θ
sin θ
5
=
3
− √334
5
3
· −√
3
34
5
sin θ = − √
34
√
5 34
sin θ = −
34
sin θ =
EX 3:
find sinθ if tanθ =
8
3
cot θ =
3
8
1 + cot2 θ = csc2 θ
9
73
1+
=
64 √64
73
csc θ =
8
√
8 73
sin θ =
73
When you simplify problems with identities always TRY to get it down to just one trig function.
EX 4:
Simplify cos θ · tan θ
= sin θ
EX 5:
Simplify csc θ · cot θ
=
cos θ
cos θ
=
2
1 − cos2 θ
sin θ
sec x + tan x
rewrite everything in terms of sine and cosine
sec x − tan x
sin x
1
sin x
1
+
÷
−
cos x cos x cos x cos x
EX 6:
Simplify
Write as one fraction (Get a common denominator if necessary)
1 + sin x 1 − sin x
÷
cos x
cos x
1 + sin x
cos x
·
cos x
1 − sin x
1 + sin x
1 − sin x
EX 6:
Verify. Like a proof so the answer is not as important as the process.
Only manipulate one side of the equation at a time
try to get the sides to look the same
Only do one step per line
sin x
1 − cos x
+
= 2 csc x
1 − cos x
sin x
get a common denominator
(1 − cos x)(1 − cos x)
sin x(sin x)
+
= 2 csc x
(1 − cos x)(sin x)
(1 − cos x)(sin x)
sin2 x + (1 − 2 cos x + cos2 x
= 2 csc x
1 − cos x)(sin x)
(sin2 x + cos2 x + 1 + 2 cos x
= 2 csc x
1 − cos x)(sin x)
2 + 2 cos x
= 2 csc x
1 − cos x)(sin x)
2(1 + cos x)
= 2 csc x
1 − cos x)(sin x)
2
= 2 csc x
(sin x)
2 csc x = 2 csc x