Download 6.5 Double Angle and Half Angle Formulas

Section 6.5 Notes Page 1
6.5 Double Angle and Half Angle Formulas
If we have either a double angle 2θ or a half angle
θ
2
then these have special formulas:
sin( 2θ ) = 2 sin θ cos θ
cos(2θ ) = cos 2 θ − sin 2 θ
cos(2θ ) = 2 cos 2 θ − 1
cos(2θ ) = 1 − 2 sin 2 θ
tan( 2θ ) =
sin
θ
cos
tan
2
θ
2
θ
2
There are three formulas for cos( 2θ )
2 tan θ
1 − tan 2 θ
=±
1 − cos θ
2
You will choose plus or minus depending on what quadrant
=±
1 + cos θ
2
You will choose plus or minus depending on what quadrant
=±
1 − cos θ
1 + cosθ
You will choose plus or minus depending on what quadrant
There are better formulas for tan
tan
θ
2
=
sin θ
,
1 + cos θ
tan
θ
2
=
θ
2
θ
2
θ
2
θ
2
is.
is.
is.
that don’t involve a plus or minus:
1 − cos θ
sin θ
EXAMPLE: Compute sin θ , tan θ , cscθ , secθ , cot θ , sin( 2θ ) , cos( 2θ ) , tan( 2θ ) , sin
θ
if you are given cosθ = −0.8 and 90 ≤ θ ≤ 180 . Round decimals to two decimal places.
o
2
, cos
θ
2
, and tan
θ
2
o
We need to draw a triangle for this one. We are given that the triangle should be drawn in the second quadrant.
− 0 .8
We can rewrite our problem as cos θ =
. We know the adjacent side is –0.8. The hypotenuse is 1. Once
1
we label our triangle we find the third side by using the Pythagorean theorem.
1
0.6
θ
-0.8
Section 6.5 Notes Page 2
Now we can get our first 5 trigonometric functions by reading off our triangle:
sin θ = 0.6
tan θ =
csc θ =
0 .6
= −0.75
− 0 .8
1
= 1.67
0 .6
cot θ =
sec θ =
1
= −1.25
− 0 .8
− 0 .8
= −1.33
0 .6
Next I will find sin( 2θ ) by using its formula: sin( 2θ ) = 2 sin θ cos θ . We already know sine and cosine, so we
will substitute in those decimals: sin( 2θ ) = 2(0.6)( −0.8) = −0.96 .
Next I will find cos( 2θ ) by using its formula: cos(2θ ) = 2 cos 2 θ − 1 . Notice I had a choice of three formulas
to use. Any of them would give you the correct answer. We already know cosθ = −0.8 , so we will substitute
in this decimal: cos(2θ ) = 2(−0.8) 2 − 1 = 0.28 .
2 tan θ
. We already know tan θ = −0.75 , so we
1 − tan 2 θ
− 1 .5
2(−0.75)
will substitute in this decimal: tan( 2θ ) =
=
= −3.43 .
2
0.4375
1 − (−0.75)
Next I will find tan( 2θ ) by using its formula: tan( 2θ ) =
For the half angle formulas I need to determine which quadrant
θ
2
is in. To do this let’s first start with our
given statement 90 o ≤ θ ≤ 180 o . If I divide everything by two we get: 45 o ≤
in the first quadrant, so sine, cosine, and tangent of
2
2
≤ 90 o . This tells us that
θ
2
is
should all be positive.
1 − cosθ
θ
. We chose a positive because is in the
2
2
2
2
second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal:
θ
1 − (−0.8)
1.8
sin =
=
= .9 = 0.95 .
2
2
2
Now I will find sin
θ
by using its formula: sin
θ
θ
θ
1 + cos θ
θ
. We chose a positive because is in the
2
2
2
2
second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal:
θ
1 + (−0.8)
0.2
cos =
=
= .1 = 0.32 .
2
2
2
Now I will find cos
θ
by using its formula: cos
=
sin θ
. We already know
2
2 1 + cos θ
θ
0.6
cosθ = −0.8 , and sin θ = 0.6 so we will substitute in these decimals: tan =
= 3.
2 1 + (−0.8)
Finally I will find tan
θ
θ
=
. I have three formulas to choose. I will choose: tan
θ
=
Section 6.5 Notes Page 3
EXAMPLE: Compute sin θ , tan θ , cscθ , secθ , cot θ , sin( 2θ ) , cos( 2θ ) , tan( 2θ ) , sin
θ
2
, cos
θ
2
, and tan
θ
2
1
if you are given cot θ = and 180 o ≤ θ ≤ 270 o .
2
We need to draw a triangle first. We are given that the triangle should be drawn in the third quadrant. We
know the adjacent side is 1 and the opposite side is 2. However because we are in the third quadrant we need to
make the 1 and 2 negative. Once we label our triangle we find the third side by using the Pythagorean theorem.
This will give us 5 .
-1
θ
-2
5
Now we can get our first 5 trigonometric functions by reading off our triangle:
sin θ = −
2
csc θ = −
5
2
5
=−
2 5
5
cos θ = −
1
5
=−
5
5
secθ = − 5
tan θ = 2
Next I will find sin( 2θ ) by using its formula: sin( 2θ ) = 2 sin θ cos θ . We already know sine and cosine, so we
⎛ 2 5 ⎞⎛
5 ⎞ 20 4
⎟=
⎟⎜ −
will substitute in those fractions: sin( 2θ ) = 2⎜⎜ −
⎟ 25 = 5 .
⎜
⎟
5
5
⎠
⎠⎝
⎝
Next I will find cos( 2θ ) by using its formula: cos(2θ ) = 2 cos 2 θ − 1 . Notice I had a choice of three formulas
to use. Any of them would give you the correct answer. We already know cos θ = −
5
, so we will substitute
5
2
⎛
5⎞
⎟ − 1 . This simplifies to: 2⎛⎜ 5 ⎞⎟ − 1 = 2 − 1 = − 3 .
in this fraction: cos(2θ ) = 2⎜⎜ −
⎟
5
5
⎝ 25 ⎠
⎝ 5 ⎠
Next I will find tan( 2θ ) by using its formula: tan( 2θ ) =
substitute in this number: tan( 2θ ) =
2( 2)
4
=− .
2
3
1 − ( 2)
2 tan θ
. We already know tan θ = 2 , so we will
1 − tan 2 θ
For the half angle formulas I need to determine which quadrant
Section 6.5 Notes Page 4
θ
2
is in. To do this let’s first start with our
given statement 180 o ≤ θ ≤ 270 o . If I divide everything by two we get: 90 o ≤
is in the second quadrant, so sine of
θ
2
θ
2
≤ 135 o . This tells us that
should be positive, and the cosine and tangent of
2
2
should be negative.
1 − cosθ
θ
. We chose a positive because is in the
2
2
2
2
second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal:
Now I will find sin
θ
by using its formula: sin
⎛
5⎞
⎟
1 − ⎜⎜ −
5 ⎟⎠
θ
⎝
sin =
=
2
2
θ
θ
θ
=
5+ 5
5+ 5
5
=
.
2
10
1 + cosθ
θ
. We chose a negative because is in the
2
2
2
2
second quadrant. We already know cosθ = −0.8 , so we will substitute in this decimal:
Now I will find cos
θ
by using its formula: cos
θ
=−
⎛
5⎞
5− 5
⎟
1 + ⎜⎜ −
⎟
5
θ
5− 5
⎝
⎠ =−
5
.
cos = −
=−
2
2
2
10
Finally I will find tan
θ
. I have three formulas to choose. I will choose: tan
θ
=
sin θ
. We already know
1 + cos θ
2
2
cosθ = −0.8 , and sin θ = 0.6 so we will substitute in these decimals:
2 5
2 5
−
−
θ
5
5 = − 2 5 ⋅ 5 = − 2 5 ⋅ 5 + 5 = − 10 5 − 10 = − 10 5 − 10 = − 5 − 1 .
tan =
=
2
5 5− 5
25 − 5
20
2
⎛
5− 5 5+ 5
5 ⎞ 5− 5
⎟
1 + ⎜⎜ −
⎟
5
⎝ 5 ⎠
EXAMPLE: Compute sin(22.5 o ) and tan(22.5 o ) using a half-angle formula.
⎛ 45 o ⎞
1 − cos 45 o
θ
⎟⎟ . Then we know that θ is 45 degrees, so now we use: sin =
We can write this as sin ⎜⎜
. It is
2
2
⎝ 2 ⎠
positive since 22.5 is in the first quadrant. You get: sin
45 o
sin 45 o
=
=
For tan(22.5 o ) we will use: tan
2
1 + cos 45 o
θ
2
2
2
=
1−
2
2
2 =
2− 2
2
=
2
2− 2
=
4
2− 2
.
2
2
2
2− 2 2 2 −2
2
=
=
⋅
=
= 2 −1.
2
2 2+ 2 2+ 2 2− 2
1+
2
2
EXAMPLE: Establish the identity:
cot θ − tan θ
= cos 2θ
cot θ + tan θ
Section 6.5 Notes Page 5
First we want to change these into sines and cosines.
cos θ sin θ
−
sin θ cos θ = cos 2θ
cos θ sin θ
+
sin θ cos θ
Now get common denominators on the top and bottom.
⎛ cos θ ⎞ cos θ sin θ ⎛ sin θ ⎞
−
⋅⎜
⎜
⎟⋅
⎟
⎝ cos θ ⎠ sin θ cos θ ⎝ sin θ ⎠ = cos 2θ
⎛ cos θ ⎞ cos θ sin θ ⎛ sin θ ⎞
+
⋅⎜
⎜
⎟⋅
⎟
⎝ cos θ ⎠ sin θ cos θ ⎝ sin θ ⎠
Multiply and write over a single denominator
cos 2 θ − sin 2 θ
cos θ sin θ = cos 2θ
cos 2 θ + sin 2 θ
cos θ sin θ
We will get rid of the double fractions and use cos 2 θ + sin 2 θ = 1
cos 2 θ − sin 2 θ = cos 2θ
The left side is the identity for cos 2θ .
cos 2θ = cos 2θ
EXAMPLE: (4 sin θ cos θ )(1 − 2 sin 2 θ ) = sin 4θ
First we will use the identity 1 − 2 cos 2 θ = cos 2θ . Now our problem becomes:
(4 sin θ cosθ ) cos 2θ = sin 4θ
We know that sin 2θ = 2 sin θ cosθ . We want to rewrite our problem as the following:
2 ⋅ 2 sin θ cosθ ⋅ cos 2θ = sin 4θ
Now we can use the identity sin 2θ = 2 sin θ cosθ .
2 sin 2θ cos 2θ = sin 4θ
We know that 2 sin 2θ cos 2θ is the same as sin (2 ⋅ 2θ ) = sin 4θ .
sin 4θ = sin 4θ