10.3 double-angle and half-angle formulas objective 1)to derive and apply double-angle and...
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10.3 Double-Angle and Half-Angle Formulas
Objective 1) To derive and apply double-angle and
half-angle formula for sine, cosine, and tangent.
2) To use the double-angle and half-angle formulas to simplify trigonometric expressions and verify the trigonometric identities.
3) To apply the double-angle and half-angle formulas in solving some trigonometric equations.
Question Does sin2 = 2sin ?
Use some specific value to verify the above equality is correct or wrong.
If take = . The above equality is correct.
If take = /2, then
sin2 = sin = 0
and
2sin = 2 sin(/2) = 2.
Therefore, sin2 2sin . What is sin2 ?
Example. An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes. A first approximation to the sawtooth curve is given by
-4 -2 2 4
-4
-2
2
4
-4 -2 2 4
-4
-2
2
4
It has the exact graph as
y = 2sin(x)cos3(x)
This example tell us that we need to explore and derive the double-angle formula
The double-angle formulas can be obtained very easily from the Sum and Difference formulas for sine and cosine. Recall
sin( ) sin cos cos sin
sin( ) sin cos cos sin
cos( ) cos cos sin sin
cos( ) cos cos sin sin
(1)
(2)
(3)
(4)
To derive the double-angle formula for sine, we just need set = in formula (1)
sin( ) sin cos cos sin
sin( ) sin cos cos sin
cos( ) cos cos sin sin
cos( ) cos cos sin sin
(1)
(2)
(3)
(4)
We get
sin2 = sin cos + cos sin = 2sin cos
sin2 = 2sin cos
To derive the double-angle formula for cosine, we just need set = in formula (3)
sin( ) sin cos cos sin
sin( ) sin cos cos sin
cos( ) cos cos sin sin
cos( ) cos cos sin sin
(1)
(2)
(3)
(4)
We getcos2 = cos2 – sin2 = (1 – sin2 ) – sin2
= 1 – 2sin2 = cos2 – ( 1 – cos2 ) = 2 cos2 – 1
cos2 = cos2 – sin2 = 1 – 2sin2 = 2cos2 – 1
2
tan tan 2 tantan 2
1 tan tan 1 tan
To derive the double-angle formula for tangent, we just need set = in tangent of sum-angle formula.
tan tantan( )
1 tan tan
2
2 tantan 2
1 tan
Summary of Double-Angle Formulas
sin2 = 2sin cos
cos2 = cos2 – sin2 = 1 – 2sin2
= 2cos2 – 1
2
2 tantan 2
1 tan
You can see from these four formulas that the double-angle formula can be interpreted as a function of same (or co)trig function of single-angle but with higher power. It seems like
F(double-angle) = G(single-angle) higher power.
Double-angle formula for cosine can be expressed in more than one way.
Example 1: If , find sin2x given that
sin x < 0.3
1cos x
2 2 1sin cos 1, cos ,
3x x x
22 1
sin 1 3
x
2 8sin
9x
9
24
3
1
3
222cossin22sin
xxx
[Solution] Recall the Pythagorean relationship
2 2Since sin 0, sin
3x x
Example 2: If , and 0 < < /2, find
sin2 , cos2, and tan2 .
4sin
5
2 2 4sin cos 1, sin , and 0 <
5 2
3 cos ,
5
4 3 24sin 2 2sin cos 2
5 5 25
sin 2 24 / 25 24tan 2
cos 2 7 / 25 7
[Solution] Recall the Pythagorean relationship
22 4 7
Since cos 2 1 2sin , cos 2 1 25 25
This is the book example on P. 381. My way is slightly different.
Example 3: Find the exact value of
(a) 2sin67o30’cos67o30’ (b) 2 25sin sin
12 12
sin 2 2sin cos
o o o o 2 2sin 67 30 'cos67 30 ' sin(2 67 30 ') sin135
2
5(b) Since sin cos cos ,
12 2 12 12
2 25 5 5 5 3cos sin cos 2 cos
12 12 12 6 2
[Solution] (a) Recall the double-angle formula for sine:
2 2 2 2
Then
5 5 5 sin sin sin cos
12 12 12 12
Example 4: Prove the identity
sin cos4 4
2sec 2sin cos
4 4
x xx
x x
sin cos sin cos cos sin4 4 4 4 4 4
sin cos sin cos4 4 4 4
x x x x x x
x x x x
sin4 4
sin cos4 4
x x
x x
[Proof] Start from the left side
2sin2
2sin cos4 4
x x
2
sin 22
x
2
cos 2x 2sec 2x
Practice: Simplify the following:
2 2 2
2 2 2
4 4
(a) cos sin (b) sec cos 2cos2 2
3(c) sin 2cos + cos (d) 2sin sin( )
2 2 2
(e) cos sin
x x x x x
x x
(a) cos(2 ) (b) cos 2
(c) cos (d) sin 2
(e) cos 2
x x
Key
Half-Angle FormulasAfter we get the double-angle formula for sine, cosine and tangent, if we make backwards substitution in cosine double-angle formulas, we can get half-angle formulas easily.
cos2 = 1 – 2sin2 = 2cos2 – 1
We let = 2, then = /2, so the above formulas are:
cos = 1 – 2sin2 /2 (1)
cos = 2cos2 /2 – 1 (2)
The above two quadratic equations are with respect to sine and cosine of /2. Solve these two equations, we get
Half-Angle FormulasAfter we get the double-angle formula for sine, cosine and tangent, if we make backwards substitution in cosine double-angle formulas, we can get half-angle formulas easily.
cos2 = 1 – 2sin2 = 2cos2 – 1
We let = 2, then = /2, so the above formulas are:
cos = 1 – 2sin2 /2 (1)
cos = 2cos2 /2 – 1 (2)
The above two quadratic equations are with respect to sine and cosine of /2. Solve these two equations, we get
22sin 1 cos 1 cos
sin2 2
2
Therefore, the tan /2 can be directly derived from half-angle of sine and cosine above:
1 cossin
221 coscos
2
1 cost
2 1 cos
2
an
In textbook, the statement on top of P. 383 addresses the alternative half-angle formulas for tangent can be derived by simplifying the radical expression of tan /2. HOW?
22cos 1 cos 1 cos
cos2 2
2
In my personal opinion, that statement is not workable. Actually, the tan /2 alternative formulas without the ambiguous sign can be derived as following:
sin 2sin cos2 2tan
22
cos 2cos cos2 2 2
Notice the Pythagorean relationship: sin2 = 1 – cos2,
2
sin
2cos2
sin
1 cos
or, sin2 = (1 – cos )(1 + cos )
Dividing sin (1 + cos ) at both sides, we obtain:sin
1 co
os
s
1 c
sin
sin 1 costan
2 1 cos sin
Or, the other alternative formula for tan /2 without the ambiguous sign can be derived as following:
sin 2sin sin2 2tan
22
cos 2cos sin2 2 2
You can see that in the trigonometry, there is more than one way to get same or an equivalent expression.
22sin2
sin
1 cos
sin
One of the alternative half-angle formulas for tan /2 can be derived in a very nice geometric way (pretended that is an acute angle):
sintan
2 1 cos
BC
AC
In right triangles BOC, OC = cos, BC = sin.
Then
y
x0 /2
1 sin
cos1A
B
CIn right triangles ABC, AC = 1 + cos.
Again!!! You can see that in the trigonometry, there is more than one way to get the same or an equivalent expression.
Summary of Half-Angle Formulas
1 cossin
2 2
1 coscos
2 2
sintan
2 1 cos
1 cos
sin
1 costan
2 1 cos
(1)
(2)
(3)
(4)
(5)
Example 5: Find the exact value of 5
cos8
5cos 0
8
51 cos5 4cos
8 2
21
2
2
[Solution] Since 5/8 is in the 2nd Quadrant,
2 2 2 2
4 2
Example 6: Find the exact value of
sin15o, cos15o, and tan15o
oo
311 cos30 2 3 2 32sin15 = = =
2 2 4 2
22( 3 1) ( 3 1) 2
16 4
6 2
4
[Solution] Apply the half-angle formula:
22(2 3) 4 2 3 3 2 3 1 ( 3 1)
8 8 8 8
Example 6: Find the exact value of
sin15o, cos15o, and tan15o
oo
311 cos30 2 3 2 32cos15 = = =
2 2 4 2
22( 3 1) ( 3 1) 2
16 4
6 2
4
[Solution] Apply the half-angle formula:
22(2 3) 4 2 3 3 2 3 1 ( 3 1)
8 8 8 8
Example 6: Find the exact value of
sin15o, cos15o, and tan15o
oo
o
1sin 30 2tan15 = =
1 cos30 31
2
[Solution] Apply the half-angle formula:
1 2 32 3
2 3 2 3
1 =
2 3
Practice: Find the exact values for:o o o(a) sin22.5 (b) cos22.5 (c) tan 22.5
2 2 2 2(a) (b) (c) 2 1
2 2
Key
Example 7: Write 1 + cos + sin as product.
21 cos 2cos ,2
[Solution] Notice the double-angle and half-angle formula:
Therefore,
sin 2sin cos2 2
21 cos sin 2cos 2sin cos2 2 2
2cos cos sin2 2 2
Example 8: If is in the 2nd quadrant, and
tan = – 4/3. Find
(a) (b) (c)
3cos = ,
5
[Solution] Since is in the 2nd quadrant, tan = – 4/3, so
4 2 2
k k
2 22
k k
sin2
cos
2
tan
2
r2 = 42 + (– 3)2 = 25, r = 5.
Therefore,
/2
y
x0–3
4 5 /2
4sin = .
5
If k is odd, then /2 is in 3rd quadrant. If k is even, then /2 is in 1st quadrant.
Example 8: If is in the 2nd quadrant, and
tan = – 4/3. Find
(a) (b) (c)
1 cos 1 ( 3 / 5)sin =
2 2 2
4 =
5
sin2
cos
2
tan
2
(a)
/2
y
x0–3
4 5 /2
[Solution] If k is odd, then /2 is in 3rd quadrant. So
2 5 =
5
1 cos 1 ( 3 / 5)cos =
2 2 2
1 =
5
(b)
5 =
5
Example 8: If is in the 2nd quadrant, and
tan = – 4/3. Find
(a) (b) (c)
sin 4 / 5tan =
2 1 cos 1 ( 3 / 5)
=2
sin2
cos
2
tan
2
(c)
/2
y
x0–3
4 5 /2
[Solution] If k is odd, then /2 is in 3rd quadrant. So
If k is even, then /2 is in 1st quadrant.
1 cos 1 ( 3 / 5)sin =
2 2 2
4 =
5
(a)
2 5 =
5
Example 8: If is in the 2nd quadrant, and
tan = – 4/3. Find
(a) (b) (c)sin2
cos
2
tan
2
/2
y
x0–3
4 5 /2
[Solution] If k is even, then /2 is in 1st quadrant.
1 cos 1 ( 3 / 5)cos =
2 2 2
1 =
5
(b)
5 =
5
sin 4 / 5tan =
2 1 cos 1 ( 3 / 5)
=2(c)
PracticeSimplify the expression by using a double-angle or half-
angle formula.
1.
2.
3.
4. Find if and u is in
quadrant IV.
2sin 35 sin 35
2 2cos sin2 2
1 cos 4
2
sin , cos , and tan ,2 2 2
u u u 2cos
3u
o(1) 1 cos70 (2) cos (3) sin 2
KeyPractice
3(4) 2 2 2 ,
2k u k k
Z
3,
4 2
uk k k
Z
is in 2nd Quad., if is even.2
uk
is in 4th Quad., if is odd.2
uk
If is even
6 30 5sin cos tan
2 6 2 6 2 5
k
u u u
KeyPractice
If is odd
6 30 5sin cos tan
2 6 2 6 2 5
k
u u u
Power-Reducing FormulaPower-Reducing Formula
From the double-angle formulas, we have learned that the expression of double-angle is ended in angle-halving and power-raising.
sin2 = 2sin cos
cos2 = cos2 – sin2 = 1 – 2sin2
= 2cos2 – 1
2
2 tantan 2
1 tan
Angle-halving2nd power
Power-Reducing FormulaPower-Reducing Formula
2
2cos1sin 2 x
x
2
2cos1cos2 x
x
x
xx
2cos1
2cos1tan2
From the half-angle formulas, we have learned that the expression of half-angle is ended in angle-doubling and power-reducing.
Angle-doubling 1st power
22 cossin3
Example 9: Reduce the power of:
[Solution 1] The expression has power of 4. Apply the power-reducing formula, we have
2 2 1 cos 2 1 cos 23sin cos 3
2 2
3(1 cos 2 ) (1 cos 2 )
4
2 23 3(1 cos 2 ) sin 2
4 4
3 1 cos 4 3(1 cos 4 )
4 2 8
22 cossin3
Example 9: Reduce the power of:
[Solution 2] The expression has power of 4. Apply the double-angle formula reversely, we have
2 2 2 233sin cos 4sin cos
4
23(2sin cos )
4
2 23 3(sin 2 ) sin 2
4 4
3 1 cos 4 3(1 cos 4 )
4 2 8
02sincos2 xxExample 10: Solve the trig equation:
Solving More Difficult Trigonometric EquationSolving More Difficult Trigonometric Equation
[Solution] Like solving any polynomial equation, it will be much easier to solve the trigonometric equation in a factor form. We start to factor the equation.
2cos sin 2 2cos 2sin cos 2cos (1 sin )x x x x x x x
Therefore, the original equation becomes:
2cos (1 sin ) 0.x x
After we learned the double-angle and half-angle and power-reducing formulas, we can solve some more difficult trigonometric equation.
So,cos 0, or 1 sin 0.x x
02sincos2 xxExample 10: Solve the trig equation:
Solving More Difficult Trigonometric EquationSolving More Difficult Trigonometric Equation
[Solution]
cos 0, , .2
x x k k Z
1 sin 0, sin 1 2 , 2
x x x k k Z
cos 0, or 1 sin 0.x x
So the solutions are
, or 2 , 2 2
x k x k k Z
2cos2sin2 22 x
x
Example 11: Solve the trig equation:
[Solution] Rewrite the expression at the left,
22cos 1 cos2
xx
2 2 22 sin 1 1 sin 1 cosx x x
Power-reducing the expression at the right,
So we end up with an trig equation:21 cos 1 cosx x cos (cos 1) 0x x
cos 0, or cos 1.x x
, or 2 (2 1) , 2
x k x k k k Z
sin 2 sin 6 0x x Example 12: Solve the trig equation:
[Solution] Rewrite sum to product
sin 4 cos( 2 ) 0x x
2 6 2 62sin cos 0
2 2
x x x x
Then
So,sin 4 0, or cos 2 0.x x
4 , or 2 , 2
x k x k k Z
sin 4 cos 2 0x x
, or , 4 2 4
k kx x k
Z
Linear Combination of Sine and CosineLinear Combination of Sine and Cosine
A linear combination of sine and cosine is always equivalent to a sine or cosine plus a non-zero phase. In the next chapter we will use this skill to switch an algebraic expression of a complex number to the polar form. In the Calculus class you will find this skill is very useful while you deal with an integral.
What is a linear combination of sine and cosine?
A linear combination of sine and cosine is of the form
a sin + b cos , where a, b R
How can we change this linear combination to a sine or cosine?
Here is what we will do:
Linear Combination of Sine and CosineLinear Combination of Sine and Cosine
Step 1 Find r: 2 2 .r a b
Step 2 Factor out r in the linear combination
sin cosa b 2 2
2 2 2 2sin cos
a ba b
a b a b
Step 3 Use the segment joining point (a, b) and the origin as terminal side, the angle formed with positive x-axis is the angle we are looking for as a phase, denoted as ϕ. Therefore,
2 2 2 2cos , and sin
a b
a b a b
Linear Combination of Sine and CosineLinear Combination of Sine and Cosine
Step 4 Rewrite the linear combination as
2 2
2 2 2 2sin csin c s so o
a ba b
aa b
b a b
where ϕ is the angle formed by segment joining point (a, b) and the origin and positive x-axis.
2 2 cos sin sin cosa b
2 2 sin cos cos sina b
2 2 sin( )a b
2 2sin cos sin( )a b a b
2 sin cos 2x x Example 13: Solve the trig equation:
[Solution] Rewrite the linear combination
2 13 sin cos 2
3 3x x
2( 2) 1 3r
Then factor out r
Let2 1
cos , and sin .3 3
3 sin cos cos sin 2.x x
3 sin( ) 2x 2 6
sin( )33
x
2 sin cos 2x x Example 13: Solve the trig equation:
[Solution]
Then the general solution for sin(x – ϕ) above is
3 sin( ) 2x 2 6
sin( )33
x
6( 1) arcsin ,
3kx k k
Z
Since ϕ can be expressed as3
arcsin ,3
6 3( 1) arcsin arcsin ,
3 3kx k k
Z
Therefore, the general solution is
Assignment
P. 383 #1 – 18, 19 – 25 (odd), 31, 33, 37
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