1. 10.3 Double-Angle and Half-Angle Formulas
Objective
To derive and apply double-angle and half-angle formula for sine, cosine, and tangent.
To use the double-angle and half-angle formulas to simplify trigonometric expressions and verify the trigonometric identities.
To apply the double-angle and half-angle formulas in solving some trigonometric equations.

2. 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 ?

3. It has the exact graph as
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
y = 2sin(x)cos3(x)
It has the exact graph as

4. 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
(1)
(2)
(3)
(4)

5. sin2 = sin cos + cos sin 
To derive the double-angle formula for sine, we just need set  =  in formula (1)
(1)
(2)
(3)
(4)
We get
sin2 = sin cos + cos sin 
= 2sin cos
sin2 = 2sin cos

6. cos2 = cos2 – sin2 = 1 – 2sin2 = 2cos2 – 1
To derive the double-angle formula for cosine, we just need set  =  in formula (3)
(1)
(2)
(3)
(4)
We get
cos2 = cos2 – sin2 = (1 – sin2 ) – sin2
= 1 – 2sin2
= cos2 – ( 1 – cos2 )
= 2 cos2 – 1
cos2 = cos2 – sin2 = 1 – 2sin2 = 2cos2 – 1

7. To derive the double-angle formula for tangent, we just need set  =  in tangent of sum-angle formula.

8. Summary of Double-Angle Formulas
sin2 = 2sin cos
cos2 = cos2 – sin2
= 1 – 2sin2
= 2cos2 – 1
Double-angle formula for cosine can be expressed in more than one way.
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.

9. Example 1: If , find sin2x given that sin x < 0.
[Solution] Recall the Pythagorean relationship

10. Example 2: If , and 0 <  < /2, find sin2 , cos2, and tan2 .
[Solution] Recall the Pythagorean relationship
This is the book example on P My way is slightly different.

11. Example 3: Find the exact value of (a) 2sin67o30’cos67o30’ (b)
[Solution] (a) Recall the double-angle formula for sine:

12. Example 4: Prove the identity
[Proof] Start from the left side

13. Practice: Simplify the following:
Key

14. Half-Angle Formulas
After 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

15. Half-Angle Formulas
After 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

16. Therefore, the tan /2 can be directly derived from half-angle of sine and cosine above:
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?

17. In my personal opinion, that statement is not workable
In my personal opinion, that statement is not workable. Actually, the tan /2 alternative formulas without the ambiguous sign  can be derived as following:
Notice the Pythagorean relationship: sin2 = 1 – cos2,
or, sin2 = (1 – cos )(1 + cos )
Dividing sin (1 + cos ) at both sides, we obtain:

18. Or, the other alternative formula for tan /2 without the ambiguous sign  can be derived as following:
You can see that in the trigonometry, there is more than one way to get same or an equivalent expression.

19. In right triangles BOC, OC = cos, BC = 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): y x
 /2  1
sin
cos A B C
In right triangles BOC, OC = cos,
BC = sin.
In right triangles ABC, AC = 1 + cos.
Then
Again!!! You can see that in the trigonometry, there is more than one way to get the same or an equivalent expression.

20. Summary of Half-Angle Formulas
(1)
(2)
(3)
(4)
(5)

21. Example 5: Find the exact value of
[Solution] Since 5/8 is in the 2nd Quadrant,

22. Example 6: Find the exact value of sin15o, cos15o, and tan15o
[Solution] Apply the half-angle formula:

23. Example 6: Find the exact value of sin15o, cos15o, and tan15o
[Solution] Apply the half-angle formula:

24. Example 6: Find the exact value of sin15o, cos15o, and tan15o
[Solution] Apply the half-angle formula:

25. Practice: Find the exact values for:
Key

26. Example 7: Write 1 + cos + sin as product.
[Solution] Notice the double-angle and half-angle formula:
Therefore,

27. Example 8: If  is in the 2nd quadrant, and tan = – 4/3. Find
(a) (b) (c)
[Solution] Since  is in the 2nd quadrant, tan = – 4/3, so
r2 = 42 + (– 3)2 = 25, r = 5.
Therefore,  y x –3 4 5
 /2
 /2
If k is odd, then  /2 is in 3rd quadrant. If k is even, then
 /2 is in 1st quadrant.

28. Example 8: If  is in the 2nd quadrant, and tan = – 4/3. Find
(a) (b) (c)
[Solution] If k is odd, then  /2 is in 3rd quadrant. So  y x –3 4 5
(a)
 /2
 /2
(b)

29. Example 8: If  is in the 2nd quadrant, and tan = – 4/3. Find
(a) (b) (c)
[Solution] If k is odd, then  /2 is in 3rd quadrant. So  y x –3 4 5
(c)
 /2
If k is even, then  /2 is in 1st quadrant.
 /2
(a)

30. Example 8: If  is in the 2nd quadrant, and tan = – 4/3. Find
(a) (b) (c)
[Solution] If k is even, then  /2 is in 1st quadrant.  y x –3 4 5
(b)
 /2
 /2
(c)

31. Practice
Simplify the expression by using a double-angle or half-angle formula. 1. 2. 3.
Find if and u is in
quadrant IV.

32. Practice
Key

33. Practice
Key

34. Power-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
Angle-halving
2nd power

35. Power-Reducing Formula
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

36. Example 9: Reduce the power of:
[Solution 1] The expression has power of 4. Apply the power-reducing formula, we have

37. Example 9: Reduce the power of:
[Solution 2] The expression has power of 4. Apply the double-angle formula reversely, we have

38. Solving More Difficult Trigonometric Equation
After we learned the double-angle and half-angle and power-reducing formulas, we can solve some more difficult trigonometric equation.
Example 10: Solve the trig 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.
Therefore, the original equation becomes:
So,

39. Solving More Difficult Trigonometric Equation
Example 10: Solve the trig equation:
[Solution]
So the solutions are

40. Example 11: Solve the trig equation:
[Solution] Rewrite the expression at the left,
Power-reducing the expression at the right,
So we end up with an trig equation:

41. Example 12: Solve the trig equation:
[Solution] Rewrite sum to product
Then
So,

42. Linear 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?

43. Linear Combination of Sine and Cosine
Here is what we will do:
Step 1 Find r:
Step 2 Factor out r in the linear combination
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,

44. Linear Combination of Sine and Cosine
Step 4 Rewrite the linear combination as
where ϕ is the angle formed by segment joining point (a, b) and the origin and positive x-axis.

45. Example 13: Solve the trig equation:
[Solution] Rewrite the linear combination
Then factor out r
Let

46. Example 13: Solve the trig equation:
[Solution]
Then the general solution for sin(x – ϕ) above is
Since ϕ can be expressed as
Therefore, the general solution is

47. Assignment
P. 383 #1 – 18, 19 – 25 (odd), 31, 33, 37