Download Double-Angle and Half-Angle Identities

Double-Angle and Half-Angle Identities
by CHED on June 16, 2017
lesson duration of 5 minutes
under Precalculus
generated on June 16, 2017 at 09:07 am
Tags: Trigonometry
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K-12 Teacher's Resource Community
Generated: Jun 16,2017 05:07 PM
Double-Angle and Half-Angle Identities
( 5 mins )
Written By: CHED on July 4, 2016
Subjects: Precalculus
Tags: Trigonometry
Resources
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Content Standard
Key concepts of circular functions, trigonometric identities, inverse trigonometric functions, and the polar coordinate
system
Performance Standard
Formulate and solve accurately situational problems involving circular functions
Apply appropriate trigonometric identities in solving situational problems
Formulate and solve accurately situational problems involving appropriate trigonometric functions
Formulate and solve accurately situational problems involving the polar coordinate system
Learning Competencies
Solve situational problems involving trigonometric identities
Prove other trigonometric identities
Simplify trigonometric expressions
Derive the double and half-angle formulas
Double-Angle Identities 1 mins
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Trigonometric identities simplify the computations of trigonometric expressions. In this lesson, we continue on
establishing more trigonometric identities. In particular, we derive the formulas for f(2)
and f 1/2
) , where f is the sine, cosine, or tangent function.
Recall the sum identities for sine and cosine.
sin(A + B) = sinAcosB + cosAsinB
cos(A + B) = cosAcosB – sinAsinB
When A = B, these identities becomes
sin 2A = sinAcosA + cosAsinA = 2sinAcosA
and
cos 2A = cosAcosA – sinAsinA = cos2 A – sin2 A.
Double-Angle Identities for Sine and Cosine
sin 2A = 2sinAcosA cos 2A = cos2 A – sin2 A
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The double-identity for cosine has other forms. We use the Pythagorean identity sin2
+ cos2
= 1.
Other Double-Angle Identities for Cosine
cos 2A = 2cos2 A – 1 cos2A = 1– 2 sin2 A
Example 3.6.1. Given sin t = 3 5 and
2<t<
, find sin2t and cos 2t.
Solution.
Solution. We first find cos t using the Pythagorean identity. Since t lies in QII, we have
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In the last example, we may compute cos 2t using one of the other two double angle identities for cosine. For the sake
of answering the curious minds, we include the computations here.
In the three cosine double-angle identities, which formula to use depends on the convenience, what is given, and what
is asked.
Example 3.6.2. Derive an identity for sin 3x in terms of sin x.
Solution. We use the sum identity for sine, the double-angle identities for sine and cosine, and the Pythagorean
identity.
sin 3x = sin(2x + x)
= sin2x cos x + cos2x sin
= (2sinx cos x) cos x + (1 – 2 sin2 x) sinx
= 2sinx cos2 x + sinx – 2 sin3 x
= 2(sinx)(1 – sin2 x) + sinx 2– sin3 x
= 3sinx – 4 sin3 x
For the double-angle formula for tangent, we recall the tangent sum identity:
tan(A + B) = tanA + tanB / 1 – tanAtanB .
When A = B, we obtain
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tan(A + A) = tanA + tanA / 1 – tanAtanA = 2 tanA / 1 – tan2 A .
Tangent Double-Angle Identity
tan 2A = 2 tanA 1 – tan2 A
Example 3.6.3. If
If tan
= –1/3 and sec
, and tan2
Solution. We can compute immediately tan 2
> 0, find sin 2
, cos2
.
.
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From the given information, we deduce that
compute cos
through sec
lies in QIV. Using one Pythagorean identity, we
. (We may also use the technique discussed in
Lesson 3.2 by solving for x, y, and r.) Then we proceed to find cos 2
.
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Seatwork/Homework 1 mins
1. If cos
= 2/ 3 and 3
, cos2
2. Express tan 3
3. Prove: 2 tan
/2 <
<2
, and tan2
.
in terms of tan
1 + tan 2
, find sin2
.
= sin2
.
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Half-Angle Identities 1 mins
Recall two of the three double-angle identities for cosine:
cos 2A = 2cos 2 A –1 and cos2A = 1– 2 sin 2 A.
From these identities, we obtain two useful identities expressing sin2 A and cos2 A in terms of cos 2A as follows:
cos 2 A = 1 + cos2A / 2 and sin 2 A = 1 – cos 2A / 2 .
Some Useful Identities
cos 2 A = 1 + cos2A / 2
sin 2 A = 1 ? cos 2A / 2
From these identities, replacing A with A/2, we get
and
These are the half-angle identities for sine and cosine.
Half-Angle Identities for Sine and Cosine
Because of the “square” in the formulas, we get
The appropriate signs of cos A/2 and sin A/2 depend on which quadrant A/2 lies.
Example 3.6.4. Find the exact values of sin 22.5º and cos 22.5º.
22.5º.
Solution.
Solution. Clearly, 22.5º lies in QI (and so sin 22.5º and cos 22.5º are both positive), and 22.5º is the half-angle of 45º.
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Example 3.6.5. Prove:
Solution.
We now derive the first version of the half-angle formula for tangent.
There is another version of the tangent half-angle formula, and we can derive it from the first version.
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Tangent Half-Angle Identities
Example 3.6.6. Find the exact value of tan
/12 .
Solution.
Example 3.6.7. If sin
= –2 5, cot
> 0, and 0 ?
<2
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, find sin
Solution. Since sin
2, cos
< 0 and cot
<3
2, and tan
> 0, we conclude the
2.
<
2 . It follows that
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/2 < 3
/4 ,
/2 <
which means that
2 lies in QII.
Seatwork/Homework 1 mins
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1. Find the exact value of tan
2. If cos
/8 .
= 3/5 and 3
/2, cos
/8 <
/2, and tan
<2
, find sin
/2 .
3. Prove: sec2 ( A/2 ) = 2 – 2 cosA / sin 2 A .
Exercises 1 mins
1. Given some information about
, find sin2?, cos2?, and tan2?.
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(a) cos
= –1 4 and
(b) sec
= 5/2 and sin
(c) tan
= –2 and 3
(d) sin
= 3/5 and tan
/2 <
<
>0
/2 <
<2
<0
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2. Given the same information as in Item (1), where 0 ?
/2, cos
<2
/2 , and tan
, find sin
/2 .
3. Express each expression as one trigonometric expression, but do not find the exact value.
4. Prove each identity.
(a) tan 2 (
2 ) = (csc? ? cot
)2
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(b) tan
(c) sec 2 (
5. If a = 2tan
/2 + cot
/2 ) = (csc 2
, express sin2
/2 = 2csc
)(2 ? 2 cos
and cos 2
)
in terms of a.
6. Find the exact value of cos 36º – cos 72º.
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7. The range R of a projectile fired at an angle
meters per second is given by
R = v 2/g sin(2
with the horizontal and with an initial velocity of v
),
where g is the acceleration due to gravity, which is 9.81 m/sec 2 near the Earth’s surface.
(a) An archer targets an object 100 meters away from her position. If she positions her arrow at an angle of 32º and
releases the arrow at the speed of 30 m/sec, will she hit her target?
(b) If sin
= 2/5, solve for v when R = 50.
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(c) Given v, what is
range?
to reach largest possible range? At this
, what is the
8. The figure shows a laser scanner projection system. The optical angle ?, throw distance D, and projected image
width W are related by the equation
Solve for W in terms D and
/2 .
9. The slope of a mountain makes an angle of 45º with the horizontal. At the base of the mountain, a cannon is fired at
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withthe
thedistance
horizontal,
where
45º<it drops on the slope of
< 90º,
and with initial
velocity
an
angle
m/sec.
Neglecting air resistance,
R (in
meters)
the mountain
from the
baseof
is v
given
by
where g is the acceleration due to gravity in m/sec 2. Express this formula for R in terms of 2
.
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