Download PRECALCULUS B TEST #5 – TRIGONOMETRIC IDENTITIES

PRECALCULUS B
TEST #5 – TRIGONOMETRIC IDENTITIES, PRACTICE
SECTION 5.1 – Fundamental Identities
5
and θ is in Quadrant IV, find the remaining trigonometric functions of θ.
13
1)
If cos θ =
2)
If tan θ = −
3)
Simplify sec θ ⋅ cot θ ⋅ csc θ . Write your answer so it does not contain a fraction.
4)
Simplify
5)
Simplify ( sec θ + 1)( sec θ − 1) .
3
and θ is in Quadrant II, find the remaining trigonometric functions of θ.
4
1 − sin 2 θ
. Write your answer so it does not contain a fraction.
1 + cot 2 θ
SECTION 5.2 – Verifying Trigonometric Identities
6)
Simplify
sec x csc x
.
+
csc x sec x
7)
Simplify cos α ( sec α + csc α ) .
8)
Simplify
cos θ
sin θ
+
.
sin θ 1 + cos θ
9)
Factor cos 2 θ − 1.
10)
Factor 3sin 2θ + 4sin θ + 1.
11)
Factor sin 4 x − cos 4 x.
12)
Verify that
tan θ
= sin θ .
sec θ
13)
Verify that sin 2 β (1 + cot 2 β ) = 1.
14)
Verify that
sin 2 θ
= sec θ − cos θ .
cos θ
15)
Verify that
16)
Verify that
1 − cos a
2
= ( cot a − csc a ) .
1 + cos a
17)
Verify that sin x − sin 3 x = sin x ⋅ cos 2 x.
1 + tan 2 θ
= tan 2 θ .
2
1 + cot θ
SECTION 5.3 – Sum and Difference Identities for Cosine
18)
Find the exact value (NO DECIMALS) of cos 195°. HINT: 195° = 225° – 30°
19)
Find the exact value (NO DECIMALS) of cos (–75°). HINT: –75° = (–30°) + (–45°)
20)
Write sin 15° in terms of its cofunction.
21)
Write cot
22)
Find an angle θ such that sin θ = cos ( 2θ − 30° ) .
23)
Find an angle θ such that cot (θ − 10° ) = tan ( 2θ + 20° ) .
24)
sin x =
25)
cos s = −
9π
in terms of its cofunction.
10
2
1
and sin y = − , x in Quadrant II and y in Quadrant IV.
3
3
Find cos ( x + y ) and cos ( x − y ) .
8
3
and sin t = − , s and t in Quadrant III. Find cos ( s + t ) and cos ( s − t ) .
17
5
SECTION 5.4 – Sum and Difference Identities for Sine and Tangent
26)
Find the exact value (NO DECIMALS) of sin 105°. HINT: 105° = 60° + 45°
27)
Find the exact value (NO DECIMALS) of tan 285°. HINT: 285° = 330° – 45°
28)
cos x = −
29)
Verify that sin ( x + y ) + sin ( x − y ) = 2sin x ⋅ cos y.
30)
Verify that
5
3
and sin y = , x and y in Quadrant II. Find sin ( x + y ) and sin ( x − y ) .
13
5
sin ( p + q )
= tan p + tan q.
cos p ⋅ cos q
SECTION 5.5 – Double-Angle Identities
31)
cos 2 22.5° − sin 2 22.5° is equivalent to what single trigonometric function?
32)
⎛π ⎞
⎛π ⎞
2sin ⎜ ⎟ ⋅ cos ⎜ ⎟ is equivalent to what single trigonometric function?
⎝ 12 ⎠
⎝ 12 ⎠
33)
2 tan 75°
is equivalent to what single trigonometric function?
1 − tan 2 75°
34)
Verify that tan A ⋅ sin 2 A = 2sin 2 A.
35)
Verify that
1 + cos 2 x
= cot x.
sin 2 x
SECTION 5.6 – Half-Angle Identities
36)
Find the exact value (NO DECIMALS) of sin 22.5°.
37)
Find the exact value (NO DECIMALS) of cos 112.5°.
38)
⎛ 7π
Find the exact value (NO DECIMALS) of tan ⎜
⎝ 8
39)
40)
41)
⎞
⎟.
⎠
1 − cos 35°
is equivalent to what single trigonometric function?
2
sin 50°
is equivalent to what single trigonometric function?
1 + cos 50°
1 + cos 24°
is equivalent to what single trigonometric function?
2
*******************ANSWERS*********************
2
1)
25
144
⎛5⎞
Use sin 2 θ + cos 2 θ = 1 to determine sin θ → sin 2 θ + ⎜ ⎟ = 1 → sin 2 θ +
= 1 → sin 2 θ =
→
169
169
⎝ 13 ⎠
144
12
12
because θ is in Quadrant IV
→ sin θ = ± → sin θ = −
169
13
13
12
−
sin θ
12
=
→ tan θ = 13 → Multiply top & bottom by 13 → tan θ = −
5
cos θ
5
13
13
equals reciprocal of sin θ → csc θ = −
12
13
equals reciprocal of cos θ → sec θ =
5
5
equals reciprocal of tan θ → cot θ = −
12
sin 2 θ =
tan θ
csc θ
sec θ
cot θ
2
2)
9
25
⎛ 3⎞
Use tan 2 θ + 1 = sec 2 θ to determine sec θ → ⎜ − ⎟ + 1 = sec 2 θ → + 1 = sec 2 θ →
= sec 2 θ →
4
16
16
⎝
⎠
25
5
5
= sec 2 θ → ± = sec θ → sec θ = − because θ is in Quadrant II
16
4
4
4
cot θ equals reciprocal of tan θ → cot θ = −
3
4
cos θ equals reciprocal of secθ → cos θ = −
5
2
16
9
⎛ 4⎞
Use sin 2 θ + cos 2 θ = 1 to determine sin θ → sin 2 θ + ⎜ − ⎟ = 1 → sin 2 θ +
= 1 → sin 2 θ =
→
25
25
⎝ 5⎠
9
3
3
→ sin θ = ± → sin θ = because θ is in Quadrant II
25
5
5
5
csc θ equals reciprocal of sin θ → csc θ =
3
sin 2 θ =
3)
4)
sec θ ⋅ cot θ ⋅ csc θ =
1 cos θ 1
1
1
1
with csc θ →
⋅
⋅
=
→ Replace
= csc 2 θ
cos θ sin θ sin θ sin 2 θ
sin θ
sin 2 θ
1 − sin 2 θ
cos 2 θ
→ Replace 1 − sin 2 θ with cos 2 θ & 1 + cot 2 θ with csc 2 θ →
→
2
1 + cot θ
csc 2 θ
1
cos 2 θ
Replace csc2 θ with
→
→ Multiply numerator by reciprocal of denominator →
1
sin 2 θ
sin 2 θ
2
2
cos θ sin θ
⋅
= cos 2 θ ⋅ sin 2 θ
1
1
5)
( sec θ + 1)( secθ − 1) = sec2 θ − secθ + sec θ − 1 = sec2 θ − 1 → Transform
tan 2 θ + 1 = sec2 θ to
tan 2 θ = sec2 θ − 1 → Replace sec 2 θ − 1 with tan 2 θ → sec 2 θ − 1 = tan 2 θ
6)
7)
8)
1
1
sec x csc x
1
1
& csc x with
+
→ Replace sec x with
→ cos x + sin x
1
1
csc x sec x
cos x
sin x
sin x cos x
1 sin x
1 cos x sin x cos x
⋅
+
⋅
=
+
→
Multiply each numerator by reciprocal of denominator →
cos x 1
sin x 1
cos x sin x
sin x ⋅ sin x cos x ⋅ cos x sin 2 x + cos 2 x
Put over a common denominator →
+
=
→
cos x ⋅ sin x cos x ⋅ sin x
cos x ⋅ sin x
1
1
1
Replace sin 2 x + cos 2 x with 1 →
→ Write as separate fractions →
⋅
→
cos x ⋅ sin x
cos x sin x
1
1
1
1
⋅
= sec x ⋅ csc x
Replace
with sec x and
with csc x →
cos x
sin x
cos x sin x
1
1
1 ⎞
⎛ 1
and csc α with
→ cos α ⎜
+
⎟→
cos α
sin α
⎝ cos α sin α ⎠
cos α
cos α
cos α
1+
with cot α → 1 +
→ Replace
= 1 + cot α
sin α
sin α
sin α
cos α ( sec α + csc α ) → Replace sec α with
cos θ ⋅ (1 + cos θ )
cos θ
sin θ
sin θ ⋅ sin θ
+
→ Put over a common denominator →
+
→
sin θ 1 + cos θ
sin θ ⋅ (1 + cos θ ) sin θ ⋅ (1 + cos θ )
cos θ + cos 2 θ
sin 2 θ
cos θ + cos 2 θ + sin 2 θ
+
→ Combine into single fraction →
→
sin θ ⋅ (1 + cos θ ) sin θ ⋅ (1 + cos θ )
sin θ ⋅ (1 + cos θ )
Replace cos 2 θ + sin 2 θ with 1 →
Reduce →
9)
cos θ + 1
1 + cos θ
→ Rearrange numerator →
→
sin θ ⋅ (1 + cos θ )
sin θ ⋅ (1 + cos θ )
1 + cos θ
1
1
1
=
→ Replace
with csc θ →
= csc θ
sin θ ⋅ (1 + cos θ ) sin θ
sin θ
sin θ
cos 2 θ − 1 → Replace cos θ with x → x 2 − 1 → Difference of Squares → x 2 − 1 = ( x + 1)( x − 1) →
Replace x with cos θ → ( cos θ + 1)( cos θ − 1)
10)
3sin 2 θ + 4sin θ + 1 → Replace sin θ with x → 3 x 2 + 4 x + 1 → Un-FOIL → 3 x 2 + 4 x + 1 = ( 3x + 1)( x + 1)
Replace x with sin θ → ( 3sin θ + 1)( sin θ + 1)
11)
sin 4 x − cos 4 x = ( sin 2 x + cos 2 x )( sin 2 x − cos 2 x ) = ( sin 2 x + cos 2 x ) ( sin x + cos x )( sin x − cos x ) →
Replace sin 2 x + cos 2 x with 1 → 1⋅ ( sin x + cos x )( sin x − cos x ) = ( sin x + cos x )( sin x − cos x )
12)
sin θ
tan θ
sin θ
1
θ →
cos
→ Replace tan θ with
→
and sec θ with
1
sec θ
cos θ
cos θ
cos θ
sin θ cos θ
sin θ cos θ
Multiply numerator by reciprocal of denominator →
⋅
→ Reduce →
⋅
= sin θ
cos θ
1
cos θ
1
13)
14)
15)
16)
⎛ cos 2 β
cos 2 β
2
sin
β
→
⎜1 +
2
sin 2 β
⎝ sin β
⎞
⎟ → Use Distributive property →
⎠
sin 2 β + cos 2 β → Replace sin 2 β + cos 2 β with 1 → sin 2 β + cos 2 β = 1
sin 2 β (1 + cot 2 β ) → Replace cot 2 β with
sin 2 θ
1 − cos 2 θ
→ Replace sin 2 θ with 1 − cos 2 θ →
→ Write as two separate fractions →
cos θ
cos θ
1
cos 2 θ
1
1
with secθ →
−
→ Simplify 2nd fraction →
− cos θ → Replace
cos θ cos θ
cos θ
cos θ
sin 2 θ
sec θ − cos θ →
= sec θ − cos θ
cos θ
1 + tan 2 θ
sec2 θ
→ Replace 1 + tan 2 θ with sec 2 θ & replace 1 + cot 2 θ with csc 2 θ →
→
2
1 + cot θ
csc 2 θ
1
2
1
1
2
2
θ →
cos
→
Change sec θ to
& csc θ to
2
2
1
cos θ
sin θ
sin 2 θ
1
sin 2 θ sin 2 θ
Multiply numerator by reciprocal of denominator →
⋅
=
→
2
1
cos θ
cos 2 θ
sin θ
sin 2 θ
1 + tan 2 θ
2
= tan θ →
=
tan
→
= tan 2 θ
θ
cos θ
cos 2 θ
1 + cot 2 θ
1 − cos a
→ Multiply numerator and denominator by conjugate of denominator →
1 + cos a
(1 − cos a ) ⋅ (1 − cos a ) 1 − cos a − cos a + cos2 a 1 − 2 cos a + cos 2 a
→
=
→
1 − cos 2 a
(1 + cos a ) ⋅ (1 − cos a ) 1 − cos a + cos a − cos2 a
Replace 1 − cos 2 a with sin 2 a →
1 − 2 cos a + cos 2 a 1 − 2 cos a + cos 2 a
=
1 − cos 2 a
sin 2 a
2
cos a
1
1 ⎞
⎛ cos a
and csc a with
→⎜
−
⎟ →
sin a
sin a
sin
sin
a
a⎠
⎝
1 ⎞ ⎛ cos a
1 ⎞ cos 2 a cos a cos a
1
⎛ cos a
Multiply with FOIL → ⎜
−
⋅
−
− 2 − 2 + 2 →
⎟ ⎜
⎟=
2
⎝ sin a sin a ⎠ ⎝ sin a sin a ⎠ sin a sin a sin a sin a
( cot a − csc a ) → Replace cot a with
2
Combine into single fraction →
cos 2 a − cos a − cos a + 1
→ Combine like terms →
sin 2 a
cos 2 a − 2 cos a + 1
→ Left-hand side now equals right-hand side
sin 2 a
17)
sin x − sin 3 x → Factor out sin x → sin x (1 − sin 2 x ) → Replace 1 − sin 2 x with cos 2 x →
sin x (1 − sin 2 x ) = sin x ⋅ cos 2 x
18)
19)
⎛
2⎞ ⎛ 3⎞ ⎛
2 ⎞ ⎛1⎞
cos195° = cos ( 225° − 30° ) = cos 225° ⋅ cos 30° + sin 225° ⋅ sin 30° = ⎜⎜ −
⎟ ⋅ ⎜⎜
⎟ + ⎜⎜ −
⎟
⎟
⎟⎟ ⋅ ⎜ ⎟ →
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝2⎠
⎛
⎛
6⎞ ⎛
2⎞
6⎞ ⎛
2⎞ − 6− 2
⎜⎜ −
⎟⎟ + ⎜⎜ −
⎟⎟ → Combine into single fraction → ⎜⎜ −
⎟⎟ + ⎜⎜ −
⎟⎟ =
4
4
4
4
4
⎝
⎠ ⎝
⎠
⎝
⎠ ⎝
⎠
cos ( −75° ) = cos ( −45° + −30° ) = cos ( −45° ) ⋅ cos ( −30° ) − sin ( −45° ) ⋅ sin ( −30° ) →
⎛ 2⎞ ⎛ 3⎞ ⎛
2 ⎞ ⎛ 1⎞
6
2
6− 2
−
→ Combine into single fraction →
⎜⎜
⎟⎟ ⋅ ⎜⎜
⎟⎟ − ⎜⎜ −
⎟⎟ ⋅ ⎜ − ⎟ =
4
4
⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2 ⎠ ⎝ 2⎠ 4
20)
21)
22)
sin θ = cos ( 90° − θ ) → sin15° = cos ( 90° − 15° ) = cos 75°
⎛π
⎞
⎛ 9π ⎞
⎛ π 9π ⎞
cot θ = tan ⎜ − θ ⎟ when θ is expressed in radians → cot ⎜
⎟ = tan ⎜ −
⎟→
⎝2
⎠
⎝ 10 ⎠
⎝ 2 10 ⎠
⎛ 5π 9π ⎞
⎛ 4π ⎞
⎛ 2π ⎞
Put over common denominator → tan ⎜
−
⎟ = tan ⎜ −
⎟ = tan ⎜ −
⎟
⎝ 10 10 ⎠
⎝ 10 ⎠
⎝ 5 ⎠
sin θ = cos ( 90° − θ ) → Replace sin θ with cos ( 90° − θ ) → sin θ = cos ( 2θ − 30° ) →
cos ( 90° − θ ) = cos ( 2θ − 30° ) → 90° − θ = 2θ − 30° → 120° = 3θ → 40° = θ
23)
cot x = tan ( 90° − x ) → Replace cot (θ − 10° ) with tan ⎡⎣90° − (θ − 10° ) ⎤⎦ →
cot (θ − 10° ) = tan ( 2θ + 20° ) → tan ⎡⎣90° − (θ − 10° ) ⎤⎦ = tan ( 2θ + 20° ) → 90° − (θ − 10° ) = 2θ + 20° →
100° − θ = 2θ + 20° → 80 = 3θ →
80°
2
= θ OR 26 ° = θ
3
3
2
24)
4
5
⎛2⎞
Use sin 2 x + cos 2 x = 1 to determine cos x → ⎜ ⎟ + cos 2 x = 1 → + cos 2 x = 1 → cos 2 x = →
3
9
9
⎝ ⎠
cos 2 x =
5
5
5
since x is in Quadrant II
→ cos x = ±
→ cos x = −
9
3
3
2
1
8
⎛ 1⎞
Use sin y + cos y = 1 to determine cos y → ⎜ − ⎟ + cos 2 y = 1 → + cos 2 y = 1 → cos 2 y = →
3
9
9
⎝
⎠
2
cos 2 y =
2
8
8
2 2
2 2
since y is in Quadrant IV
→ cos y = ±
=±
→ cos y =
9
3
3
3
⎛
5 ⎞ ⎛ 2 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 2 10 ⎞ ⎛ 2 ⎞
cos ( x + y ) = cos x ⋅ cos y − sin x ⋅ sin y = ⎜⎜ −
⎟⎟ ⋅ ⎜⎜
⎟⎟ − ⎜ ⎟ ⋅ ⎜ − ⎟ = ⎜⎜ −
⎟−⎜− ⎟ →
9 ⎟⎠ ⎝ 9 ⎠
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝
⎛ 2 10 ⎞ ⎛ 2 ⎞
−2 10 + 2
⎜⎜ −
⎟⎟ + ⎜ ⎟ → Combine into single fraction → cos ( x + y ) =
9 ⎠ ⎝9⎠
9
⎝
⎛
5 ⎞ ⎛ 2 2 ⎞ ⎛ 2 ⎞ ⎛ 1 ⎞ ⎛ 2 10 ⎞ ⎛ 2 ⎞
cos ( x − y ) = cos x ⋅ cos y + sin x ⋅ sin y = ⎜⎜ −
⎟⎟ ⋅ ⎜⎜
⎟⎟ + ⎜ ⎟ ⋅ ⎜ − ⎟ = ⎜⎜ −
⎟+⎜− ⎟ →
9 ⎟⎠ ⎝ 9 ⎠
⎝ 3 ⎠ ⎝ 3 ⎠ ⎝ 3⎠ ⎝ 3⎠ ⎝
Combine into single fraction →
−2 10 − 2
−2 10 − 2
→ cos ( x − y ) =
9
9
2
25)
64
225
⎛ 8⎞
Use sin 2 s + cos 2 s = 1 to determine sin s → sin 2 s + ⎜ − ⎟ = 1 → sin 2 s +
= 1 → sin 2 s =
→
17
289
289
⎝
⎠
sin 2 s =
225
15
15
since s is in Quadrant III
→ sin s = ± → sin s = −
289
17
17
2
9
16
⎛ 3⎞
Use sin t + cos t = 1 to determine cos t → ⎜ − ⎟ + cos 2 t = 1 →
+ cos 2 t = 1 → cos 2 t =
→
5
25
25
⎝
⎠
2
2
16
4
4
→ cos t = ± → cos t = − since t is in Quadrant III
25
5
5
13
⎛ 8 ⎞ ⎛ 4 ⎞ ⎛ 15 ⎞ ⎛ 3 ⎞ ⎛ 32 ⎞ ⎛ 45 ⎞
cos ( s + t ) = cos s ⋅ cos t − sin s ⋅ sin t = ⎜ − ⎟ ⋅ ⎜ − ⎟ − ⎜ − ⎟ ⋅ ⎜ − ⎟ = ⎜ ⎟ − ⎜ ⎟ = −
17
5
17
5
85
85
85
⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝
⎠ ⎝ ⎠ ⎝ ⎠
⎛ 8 ⎞ ⎛ 4 ⎞ ⎛ 15 ⎞ ⎛ 3 ⎞ ⎛ 32 ⎞ ⎛ 45 ⎞ 77
cos ( s − t ) = cos s ⋅ cos t + sin s ⋅ sin t = ⎜ − ⎟ ⋅ ⎜ − ⎟ − ⎜ − ⎟ ⋅ ⎜ − ⎟ = ⎜ ⎟ + ⎜ ⎟ =
⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 17 ⎠ ⎝ 5 ⎠ ⎝ 85 ⎠ ⎝ 85 ⎠ 85
cos 2 t =
26)
sin105° = sin ( 60° + 45° ) = sin 60° ⋅ cos 45° + cos 60° ⋅ sin 45° =
Combine into single fraction →
27)
3 2 1 2
6
2
⋅
+ ⋅
=
+
→
2 2 2 2
4
4
6+ 2
4
tan 330° − tan 45°
=
tan 285° = tan ( 330° − 45° ) =
1 + tan 330° ⋅ tan 45°
3
3
−1
−
−1
3
3
=
→
⎛
3
3⎞
1−
1 + ⎜⎜ −
⎟⎟ ⋅1
3
⎝ 3 ⎠
−
Multiply each term in numerator and denominator by 3 →
− 3 −3
3− 3
→
(
)
Multiply numerator and denominator by conjugate of denominator 3 + 3 →
−3 3 − 3 − 9 − 3 3
9+3 3 −3 3 −3
=
− 3 −3 3+ 3
⋅
→
3− 3 3+ 3
−12 − 6 3
−12 − 6 3
→ Reduce →
= −2 − 3
6
6
2
28)
25
144
⎛ 5⎞
Use sin 2 x + cos 2 x = 1 to determine sin x → sin 2 x + ⎜ − ⎟ = 1 → sin 2 x +
= 1 → sin 2 x =
→
169
169
⎝ 13 ⎠
sin 2 x =
144
12
12
since x is in Quadrant II
→ sin x = ± → sin x =
169
13
13
2
9
16
⎛3⎞
Use sin 2 y + cos 2 y = 1 to determine cos y → ⎜ ⎟ + cos 2 y = 1 →
+ cos 2 y = 1 → cos 2 y =
→
25
25
⎝5⎠
16
4
4
→ cos y = ± → cos y = − since y is in Quadrant II
25
5
5
12 ⎛ 4 ⎞ ⎛ 5 ⎞ 3
48 ⎛ 3 ⎞
63
sin ( x + y ) = sin x ⋅ cos y + cos x ⋅ sin y = ⋅ ⎜ − ⎟ + ⎜ − ⎟ ⋅ = − + ⎜ − ⎟ = −
13 ⎝ 5 ⎠ ⎝ 13 ⎠ 5
65 ⎝ 13 ⎠
65
12 ⎛ 4 ⎞ ⎛ 5 ⎞ 3
48 ⎛ 3 ⎞
33
sin ( x − y ) = sin x ⋅ cos y − cos x ⋅ sin y = ⋅ ⎜ − ⎟ − ⎜ − ⎟ ⋅ = − − ⎜ − ⎟ = −
13 ⎝ 5 ⎠ ⎝ 13 ⎠ 5
65 ⎝ 13 ⎠
65
cos 2 y =
29)
sin ( x + y ) + sin ( x − y ) → Replace sin ( x + y ) with sin x ⋅ cos y + cos x ⋅ sin y →
Replace sin ( x − y ) with sin x ⋅ cos y − cos x ⋅ sin y →
sin ( x + y ) + sin ( x − y ) = sin x ⋅ cos y + cos x ⋅ sin y + sin x ⋅ cos y − cos x ⋅ sin y → Combine like terms →
sin x ⋅ cos y + cos x ⋅ sin y + sin x ⋅ cos y − cos x ⋅ sin y = 2 ⋅ sin x ⋅ cos y
30)
sin ( p + q )
cos p ⋅ cos q
Write as separate fractions →
Replace
sin p ⋅ cos q + cos p ⋅ sin q
→
cos p ⋅ cos q
→ Replace sin ( p + q ) with sin p ⋅ cos q + cos p ⋅ sin q →
sin p ⋅ cos q cos p ⋅ sin q
sin p sin q
+
→ Reduce →
+
→
cos p ⋅ cos q cos p ⋅ cos q
cos p cos q
sin p
sin q
sin p sin q
with tan p and
with tan q →
+
= tan p + tan q
cos p
cos q
cos p cos q
31)
cos 2 A = cos 2 A − sin 2 A → cos 2 22.5° − sin 2 22.5° = cos ( 2 ⋅ 22.5° ) = cos 45°
32)
⎛π ⎞
⎛π ⎞
⎛ π ⎞
⎛π ⎞
sin 2 A = 2sin A ⋅ cos A → 2sin ⎜ ⎟ ⋅ cos ⎜ ⎟ = sin ⎜ 2 ⋅ ⎟ = sin ⎜ ⎟
⎝ 12 ⎠
⎝ 12 ⎠
⎝ 12 ⎠
⎝6⎠
33)
tan 2 A =
34)
tan A ⋅ sin 2 A → Replace tan A with
2 tan A
2 tan 75°
→
= tan ( 2 ⋅ 75° ) = tan150°
2
1 − tan A 1 − tan 2 75°
Simplify →
35)
sin A
sin A
and sin 2 A with 2 ⋅ sin A ⋅ cos A →
⋅ ( 2 ⋅ sin A ⋅ cos A) →
cos A
cos A
sin A
⋅ ( 2 ⋅ sin A ⋅ cos A ) = 2 ⋅ sin A ⋅ sin A = 2sin 2 A
cos A
1 + cos 2 x
→ Replace cos 2 x with ( cos 2 x − sin 2 x ) and sin 2 x with ( 2sin x cos x ) →
sin 2 x
1 + cos 2 x − (1 − cos 2 x )
1 + cos 2 x − sin 2 x
2
2
→ Replace sin x with (1 − cos x ) →
→ Simplify →
2sin x cos x
2sin x cos x
1 + cos 2 x − 1 + cos 2 x
2 cos 2 x
cos x
cos x
cos x
=
→ Reduce →
→ Replace
with cot x →
= cot x
2sin x cos x
2sin x cos x
sin x
sin x
sin x
36)
1 − cos A
1 − cos 45°
⎛ A⎞
⎛ 45° ⎞
sin ⎜ ⎟ = ±
→ sin 22.5° = sin ⎜
=±
⎟=±
2
2
⎝2⎠
⎝ 2 ⎠
Multiply each term under radical sign by 2 → ±
sin 22.5° =
1−
2
2
2 →
2− 2
2− 2
2− 2
→ Simplify → ±
=±
4
2
4
2− 2
because 22.5° is in Quadrant I
2
37)
⎛
2⎞
1 + ⎜⎜ −
⎟
2 ⎟⎠
1 + cos A
1 + cos 225°
⎛ A⎞
⎛ 225° ⎞
⎝
cos ⎜ ⎟ = ±
→ cos112.5° = cos ⎜
=±
→
⎟=±
2
2
2
⎝2⎠
⎝ 2 ⎠
Multiply each term under radical sign by 2 → ±
cos112.5° = −
38)
2− 2
2− 2
2− 2
→ Simplify → ±
=±
4
2
4
2− 2
because 112.5° is in Quadrant II
2
1 − cos A
⎛ A⎞
⎛ 7π
tan ⎜ ⎟ = ±
→ tan ⎜
1 + cos A
⎝2⎠
⎝ 8
⎛ 7π
1 − cos ⎜
⎞
⎛ 7π /4 ⎞
⎝ 4
⎟ = tan ⎜
⎟=±
⎛ 7π
⎠
⎝ 2 ⎠
1 + cos ⎜
⎝ 4
Multiply each term under radical sign by 2 → ±
(
)
conjugate of denominator 2 − 2 → ±
⎛ 7π
Reduce → ± 3 − 2 2 → tan ⎜
⎝ 4
2− 2
2+ 2
⎞
2
1−
⎟
⎠ =±
2 →
⎞
2
1+
⎟
⎠
2
→ Multiply numerator and denominator by
(2 − 2 ) ⋅(2 − 2 ) = ±
(2 + 2 ) ⋅(2 − 2 )
4−2 2 −2 2 +2
4−2 2 +2 2 −2
7π
⎞
is in Quadrant IV
⎟ = − 3 − 2 2 because
4
⎠
39)
1 − cos A
1 − cos 35°
⎛ A⎞
⎛ 35° ⎞
sin ⎜ ⎟ =
→
= sin ⎜
⎟ = sin17.5°
2
2
⎝2⎠
⎝ 2 ⎠
40)
sin A
sin 50°
⎛ A⎞
⎛ 50° ⎞
tan ⎜ ⎟ =
→
= tan ⎜
⎟ = tan 25°
⎝ 2 ⎠ 1 + cos A 1 + cos 50°
⎝ 2 ⎠
41)
1 + cos A
1 + cos 24°
⎛ A⎞
⎛ 24° ⎞
cos ⎜ ⎟ =
→
= cos ⎜
⎟ = cos12°
2
2
⎝2⎠
⎝ 2 ⎠
=±
6−4 2
2
Pythagorean Identities
Negative-Angle Identities
1)
1) sin ( −θ ) = − sin θ
2) tan 2 θ + 1 = sec 2 θ
2) cos ( −θ ) = cos θ
3) 1 + cot 2 θ = csc2 θ
3) tan ( −θ ) = − tan θ
Sum/Difference Identities
Cofunction Identities
1) cos ( A + B ) = cos A ⋅ cos B − sin A ⋅ sin B
1) sin θ = cos ( 90° − θ )
2) cos ( A − B ) = cos A ⋅ cos B + sin A ⋅ sin B
2) sec θ = csc ( 90° − θ )
3) sin ( A + B ) = sin A ⋅ cos B + cos A ⋅ sin B
3) tan θ = cot ( 90° − θ )
4) sin ( A − B ) = sin A ⋅ cos B − cos A ⋅ sin B
tan A + tan B
1 − tan A ⋅ tan B
tan A − tan B
6) tan ( A − B ) =
1 + tan A ⋅ tan B
5) tan ( A + B ) =
Double-Angle Identities
Half-Angle Identities
1) cos 2 A = cos 2 A − sin 2 A
1) cos
A
1 + cos A
=±
2
2
2) cos 2 A = 1 − 2sin 2 A
2) sin
A
1 − cos A
=±
2
2
3) cos 2 A = 2 cos 2 A − 1
3) tan
4) sin 2 A = 2 ⋅ sin A ⋅ cos A
5) tan 2 A =
2 tan A
1 − tan 2 A
A
1 − cos A
=±
2
1 + cos A
A
sin A
4) tan =
2 1 + cos A
A 1 − cos A
5) tan =
2
sin A