Section 7-4
Double-Angle and Half Angle
Identities
It is sometimes useful to have identities to find
the value of a function of twice and angle or
half an angle.
So we learned from the video that we do not need to memorize the double angle
formulas. We can derive them using the sum identity formulas!
Now we are going to use the double angle formulas to find the exact value of some
functions.
Suppose sin θ = ¾ and θ has its terminal side in the first quadrant. Find sin 2θ and
Cos 4θ
First step: Find Cos θ
16=9+7 so x =
7
Use the Pythagorean Theorem y=3 r=4 x=?
and Cos θ=
Sin 2θ= 2 sin θ cos θ = -2 (3/4)(
7
4
7
4
)= 3 7
8
Since 4θ= 2(2θ) then Cos 4θ = Cos 2(2θ)= Cos22θ‐sin22θ= 4/256‐63/64=4/256‐
252/256= ‐ 248/256 = ‐ 31/32
(Cos 2θ= Cos2θ‐sin2θ = 7/16‐9/16 = ‐2/16)
Remember when you are applying the identities,
always determine the quadrant that the
terminal side of the angle lies so that the correct
sign can be chosen.
π
7π
12
Example sin
is found in Quadrant II so the sin
will be positive.
Use a half angle identity to find the exact value of tan 15 degrees.
15 degrees= 30/2
30 degrees is a special degree.
So tan 30/2 = ±
1− cos30 =
1+ cos30
1 −
3
2
1 +
3
2
=
2 −
3
2 +
3
=
=
2 −
3
2 +
3
2 −
3
= 2 −
1
3
It will be positive because 30 degrees is in the first quadrant.
⋅
2 −
3
2 −
3
Verify that 1‐cos2x sec2x = tan2x
Š
1 ‐ 1‐2sin2x/cos2 x ? tan2x
Š
(Cos2x‐ (1‐2sin2x))/cos2 x ? tan2x Š
((Cos2 x ‐ 1) + 2sin2x)/cos2x ? tan2x
Š
‐sin2x+2sin2x/cos2x ? tan2x
Š
Sin2x/cos2x ? tan2x
Š
tan2x=tan2x
Š
Š
Š
Section 7-4
Pp. 454-455
#14-20, 28-31, 42