HW 7 Solutions
Section 6.1
α sec α
14. tan α = coscot
α .
Proof: Work with right hand side:
cos α sec α
= (cos α)(sec α)(tan α) = (cos α)
cot α
1
cos α
(tan α) = tan α.
24. (1 − sin t)(1 + sin t) = cos2 t.
Proof. Work with the left hand side:
(1 − sin t)(1 + sin t)
= 1 − sin2 t = cos2 t.
algebra
Pyth
26. (sin x + cos x)2 = 1 + 2 sin x cos x.
Proof. Work with the left hand side:
(sin x + cos x)2
= sin2 x + 2 sin x cos x + cos2 x
algebra
= (sin2 x + cos2 x) + 2 sin x cos x = 1 + 2 sin x cos x.
Pyth
30. sec2 u − tan2 u = 1.
Proof. Work with the left hand side:
sec2 u − tan2 u =
=
sin2 x
1
−
2
cos x cos2 x
1 − sin2 x
cos2 x
=
= 1.
2
cos x Pyth cos2 x
40. 1 + sin x 6= tan x + sec x.
Solution. Need to find an x for which these disagree. For instance, x =
√
left, but 2 on the right.
64. sin θ + cos θ =
tan θ+1
sec θ .
π
4
gives
Proof. Work with right hand side:
tan θ + 1
sin θ
=
+ 1 cos θ = sin θ + cos θ.
sec θ
cos θ
68. sec2 x + csc2 x = sec2 x csc2 x.
Proof. Work with the left hand side:
sec2 x + csc2 x =
1
1
sin2 x + cos2 x
+
=
cos2 x sin2 x commom denom cos2 x sin2 x
1
=
= sec2 x csc2 x.
2
2
Pyth cos x sin x
1
√1
2
on the
Section 6.2
8. (x − y)3 6= x3 − y 3 .
Solution. We need to find x and y for which the two sides disagree. Take x = 2, y = 1.
Then left side is 1 but right side is 7.
14. sin (x − y) 6= sin x−sin y. Solution. We need to find x and y for which the two sides disagree.
Take x = π, y = π2 . Then left side is sin π2 = 1 but right side is sin π − sin π2 = 0 − 1 = −1.
26. sec ( π2 − x) = csc x.
Proof. Work with the left hand side:
sec (
1
π
cofunction 1
− x) =
=
= csc x.
π
2
cos ( 2 − x)
sin x
34.
cos 105◦ = cos (60◦ + 45◦ )
√
√
3 1
1− 3
1 1
√ =
√ .
= cos (60 ) cos (45 ) − sin (60 ) sin (45 ) = √ −
2 2
2 2
2 2
◦
◦
◦
◦
36.
sin 165◦ = sin (135◦ + 30◦ )
√
√
3 1
−1 1
3+1
√ −
√ =
√ .
= cos (135◦ ) sin (30◦ ) − sin (135◦ ) cos (30◦ ) =
2 2
2 2
2 2
50. sin (2x) = 2 sin x cos x.
Proof. Plug in α = β = x into sine sum identity.
sin (2x) = sin x cos x + sin x cos x = 2 sin x cos x.
x cot y+1
52. cot (x − y) = cot
cot y−cot x .
Proof. Work with the left hand side:
cot (x − y) =
cos (x − y)
cos x cos y + sin x sin y
=
.
sin (x − y)
sin x cos y − cos x sin y
Now try the right hand side:
cot x cot y + 1
=
cot y − cot x
cos x cos y
sin x sin y + 1
cos y
cos x
sin y − sin x
The two sides are each equal to
=
cos x cos y+sin x sin y
cos x cos y
sin x cos y−cos x sin y
cos x cos y
cos x cos y+sin x sin y
sin x cos y−cos x sin y ,
=
cos x cos y + sin x sin y
.
sin x cos y − cos x sin y
which proves the desired identity.
sin (x−y)
58. tan x − tan y = cos
x cos y .
Proof. Work with the left hand side:
tan x − tan y =
sin x
sin y
sin x cos y − cos x sin x
sin (x − y)
−
=
=
cos x cos y
cos x cos y
cos x cos y
2
−1 4
−1 4
−1 4
76. cos (sin−1 ( −3
( 5 )) = cos (sin−1 ( −3
( 5 ))−sin (sin−1 ( −3
( 5 )).
5 ) + cos
5 )) cos (cos
5 )) sin (cos
−1 −3
−1
4
4
4
−3
−3
4
But cos (sin ( 5 )) = 5 , cos (cos−1 ( 5 )) = 5 , sin (sin ( 5 )) = 5 , and sin (cos−1 ( 5 )) =
−1 −3
3
3
( 5 ) + cos−1 ( 54 )) = 45 45 − −3
5 . So cos (sin
5 5 = 1.
Section 6.3
√
√
1+ 23
1−cos 150◦
= 2 + 3.
=
=
1
◦
sin 150
2
tan half angle
q √
q
√ √
1+cos 7π
1+ 22
2
1 7π
7π
4
=±
= − 2+
. (Pick negative since
20. cos ( 8 ) = cos ( 2 4 ) = ±
2
2
2
second quadrant angle.
14. tan (75◦ ) = tan ( 12 150◦ )
7π
8
is a
30. sin (2x) = tan x(1 + cos (2x)).
Proof. Work with the right side:
(tan x)(1+cos (2x)) = (
sin x
sin x
)(1+2 cos2 x−1) = (
)(2 cos2 x) = 2 sin x cos x = sin (2x).
cos x
cos x
34. 1 + sin (2t) = (sin t + cos t)2 .
Proof. Work with the right hand side:
(sin t + cos t)2 = sin2 t + 2 sin t cos t + cos2 t = (sin2 t + cos2 t) + 2 sin t cos t
=
1 + sin (2t).
Pyth and double angle
x
.
38. cot (2x) = cot x−tan
2
Proof. Work with the right hand side:
1 cos x
sin x
cos2 x − sin2 x
1
(cot x − tan x) = (
−
)=
2
2 sin x
cos x
2 sin x cos x
cos (2x)
=
= cot (2x).
double angles sin (2x)
42.
cos (2u)
1−sin (2u)
=
1+tan u
1−tan u .
Proof. Start on the left side:
cos2 u − sin2 u
cos2 u − sin2 u
cos (2u)
=
= =
1 − sin (2u)
1 − 2 sin u cos u Pyth cos2 u + sin2 u − 2 sin u cos u
=
algebra
(cos u − sin u)(cos u + sin u)
cos u + sin u
=
.
(cos u − sin u)2
cos u − sin u
Now try the right side:
1+
1 + tan u
=
1 − tan u
1−
sin u
cos u
sin u
cos u
=
cos u+sin u
cos u
cos u−sin u
cos u
=
cos u + sin u
.
cos u − sin u
Both sides of the desired identity are equal to this common thing, so the identity is proved.
3
2
sec x
44. sec (2x) = 2−sec
2 x.
Proof. Start on the right side.
1
1
sec2 x
2
=
= cos x1
2
2 − sec x
2 − cos2 x algebra 2 cos2 x − 1
1
=
= sec (2x).
double angle cos 2x
48. cos (2x) = 2 cos x?
Solution. Plug in x = 2π. Left side is cos (4π) = 1. Right side is 2 cos (2π) = 2. So this is
not an identity.
88. sin (3x) = 3 sin x − 4 sin3 x.
Proof. Work on the left hand side.
sin (3x) = sin (2x + x)
=
sin (2x) cos (x) + sin (x) cos (2x)
sine addition
=
(2 sin x cos x)(cos x) + (sin x)(cos2 x − sin2 x)
sine, cosine double angle
= 2 sin x cos2 x + sin x cos2 x − sin3 x = 3 sin x cos2 x − sin3 x
= 3(sin x)(1 − sin2 x) − sin3 x = 3 sin x − 4 sin3 x.
104. From the sketch, know tan θ =
2
x
and tan (2θ) = x6 . Using the hint, we get
4
6
x
=
.
x
1 − x42
Multiplying out denominators,
6(1 −
4
) = 4.
x2
Multiply by x2 , distribute and collect like terms, find x2 = 12 so x =
2
1
θ = arctan ( √ ) = arctan √ = 30◦ .
x 3
3
4
√
√
12 = 2 3. Then