3. Double and Half-Angle Identities
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Chapter 7
3.
Double and Half-Angle Identities
This lesson delves into the intricacies of trigonometric identities, focusing primarily on double-angle and half-angle identities. These mathematical tools are essential for simplifying complex trigonometric equations and are widely used in various fields such as engineering, physics, and computer science. Understanding these identities can make it easier to solve problems that involve angles and lengths in triangles. For instance, they are often used in calculating distances, angles, and other dimensions in real-world applications like satellite positioning and architectural design. The lesson provides a step-by-step approach to mastering these identities, making it easier for both students and professionals to apply them in practical scenarios.
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Lesson Settings & Tools
| | 12 Theory slides |
| | 9 Exercises - Grade E - A |
| | Each lesson is meant to take 1-2 classroom sessions |
Double and Half-Angle Identities
Slide of 12
This lesson will focus on introducing and practicing the trigonometric identities that relate the trigonometric values of an angle to the trigonometric values of the double-angle and half-angle.
Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Challenge
Rules of the Quest Game and a Bonus
Zosia and her classmates entered a mathematical quest game. In this quest, there is an old leather map showing the path to various sacred mathematical sites. At each site, it is either solve the problem and move forward, or suffer the consequences of the unknown of the math universe. 
At the very beginning of the quest, they were given a bonus task, which can be solved at the end of the quest. If solved successfully, they will get a huge amount of bonus points. What is the value of HD?
External credits: @brgfx
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Discussion
Presenting Double-Angle Identities
To be able to solve the tasks of the quest, the class first stopped at the infopoint to recall the Double-Angle Identities. These identities relate the trigonometric values of an angle to the trigonometric values of twice that angle.
Rule
Double-Angle Identities
The double-angle identities materialize when two angles with the same measure are substituted into the angle sum identities.
cl Sine: & sin 2θ=2sin θ cos θ [0.7em] & cos 2θ =cos^2 θ - sin^2 θ Cosine: & cos 2θ = 2cos^2 θ - 1 & cos 2θ = 1 - 2sin^2 θ [0.2cm] Tangent: & tan 2θ =2tanθ/1-tan^2 θ
Proof
Double-Angle IdentitiesStart by writing the Angle Sum Identity for sine and cosine. sin(x + y) = sin x cos y + cos x sin y cos(x + y) = cos x cos y - sin x sin y
Let x=θ and y=θ. With this, x+y becomes 2θ. Then, these two formulas can be rewritten in terms of θ. sin 2θ = sinθ cosθ + cosθ sinθ cos 2θ = cosθ cosθ - sinθ sinθ
Sine Identity
Approaching the first equation, the Commutative Property of Multiplication can be applied to the second term of its right-hand side. Then, by adding the terms on the right-hand side of this equation, the formula for sin 2θ is obtained.
sin 2θ = sinθ cosθ + sinθ cosθ ⇓ sin 2θ = 2sinθ cosθ ✓
Cosine Identities
Approaching the second equation, the Product of Powers Property can be used to rewrite its right-hand side. By doing this, the first identity for the cosine of the double of an angle is obtained.
cos 2θ = cosθ cosθ - sinθ sinθ ⇓ cos 2θ = cos^2θ - sin^2θ ✓
cos 2θ = cos^2θ - sin^2θ
▼
Substitute 1-cos^2 θ for sin^2 θ and simplify
Substitute
sin^2 θ= 1-cos^2 θ
cos 2θ = cos^2θ - ( 1-cos^2 θ)
Distr
Distribute -1
cos 2θ = cos^2θ - 1 + cos^2 θ
AddTerms
Add terms
cos 2θ = 2cos^2θ - 1 ✓
cos 2θ = cos^2θ - sin^2θ
▼
Substitute 1-sin^2 θ for cos^2 θ and simplify
cos 2θ = 1-2sin^2 θ ✓
Tangent Identity
To prove the tangent identity, start by rewriting tan 2θ in terms of sine and cosine. tan 2θ = sin 2θ/cos 2θ Next, substitute the first sine identity in the numerator and the first cosine identity in the denominator. tan 2θ = 2sin θ cos θ/cos^2 θ - sin^2 θ Then, divide the numerator and denominator by cos^2θ. tan 2θ = 2sin θ cos θcos^2θ/cos^2 θ - sin^2 θcos^2θ Finally, simplifying the right-hand side the tangent identity will be obtained.tan 2θ = 2sin θ cos θcos^2θ/cos^2 θ - sin^2 θcos^2θ
▼
Simplify right-hand side
WriteDiffFrac
Write as a difference of fractions
tan 2θ = 2sin θ cos θcos^2θ/cos^2 θcos^2θ - sin^2 θcos^2θ
CrossCommonFac
Cross out common factors
tan 2θ = 2sin θ cos θcos^2θ/cos^2 θcos^2θ - sin^2 θcos^2θ
CancelCommonFac
Cancel out common factors
tan 2θ = 2sin θcosθ/1 - sin^2 θcos^2θ
a^m/b^m=(a/b)^m
tan 2θ = 2sin θcosθ/1 - ( sin θcosθ)^2
MovePartNumLeft
a* b/c=a*b/c
tan 2θ = 2* sin θcosθ/1 - ( sin θcosθ)^2
sin(θ)/cos(θ)=tan(θ)
tan 2θ = 2tanθ/1-tan^2θ ✓
Extra
Calculating cos 120^(∘)To calculate the exact value of cos 120^(∘), these steps can be followed.
- To be able to use the double-angle identities, the angle 120^(∘) needs to be rewritten as 2 multiplied by another angle. Therefore, rewrite 120^(∘) as 2* 60^(∘).
- Use the second formula for the cosine of twice an angle.
- Based on the trigonometric ratios of common angles, it is known that cos 60^(∘) = 12.
cos 120^(∘)
Rewrite
Rewrite 120^(∘) as 2* 60^(∘)
cos (2* 60^(∘))
cos 2θ = 2cos^2 θ - 1
2cos^2 (60^(∘)) - 1
cos 60^(∘) = 1/2
2(1/2)^2 - 1
-1/2
Example
Searching for Clues
When the class got to the first mathematical sacred site, they were told to look for three clues. After eagerly searching the neighborhood, they found three parts to one task.
The task says to calculate sin 2θ knowing that cosθ= 25 and 0^(∘)≤θ≤ 90^(∘). What exact value should the class get to be able to move on to the next site?
External credits: @brgfx
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Hint
Find the value of sinθ by using one of the Pythagorean Identities.
Solution
In order to find the values of sin 2θ, recall the Double-Angle Identity for sine.
sin 2θ = 2sinθcosθ
The value of cosθ is known, but the value of sinθ is not. To find it, one of the Pythagorean Identities can be used. sin^2 θ+cos^2 θ=1
Substitute cosθ with 25 and solve for sinθ.
By the definition of absolute value, sin θ can have two possible values.
|sinθ|=sqrt(21)/5 ⇓ sinθ=sqrt(21)/5 sinθ=- sqrt(21)/5
It is given that θ is between 0^(∘) and 90^(∘), which is the first quadrant. Sine has positive values in the first quadrant, so the negative value can be disregarded.
sinθ=sqrt(21)/5
Now that both sine and cosine of θ are known, the value of sin2θ can be calculated.
Therefore, the value of sin 2θ is 4sqrt(21)25.
sin^2 θ+cos^2 θ=1
Substitute
cosθ= 2/5
sin^2 θ+( 2/5)^2=1
PowQuot
(a/b)^m=a^m/b^m
sin^2 θ+2^2/5^2=1
CalcPow
Calculate power
sin^2 θ+4/25=1
SubEqn
LHS-4/25=RHS-4/25
sin^2 θ=1-4/25
OneToFrac
Rewrite 1 as 25/25
sin^2 θ=25/25-4/25
SubFrac
Subtract fractions
sin^2 θ=21/25
|sinθ|=sqrt(21)/5
sin2θ=2sinθcosθ
SubstituteII
sinθ= sqrt(21)/5, cosθ= 2/5
sin2θ=2( sqrt(21)/5)( 2/5)
MultFrac
Multiply fractions
sin2θ=2(2sqrt(21)/25)
MoveLeftFacToNum
a*b/c= a* b/c
sin2θ=4sqrt(21)/25
Example
Finding Trigonometric Values to Determine the Password
At the second site, a steel safe awaits them. It is locked! Inside is the paper with the information needed to get to the third site. To open the safe, they must figure out the password.
The password is decoded sequentially by the integers that appear in the answers to the three given tasks. It is known that sinθ= 13 and 90^(∘)≤ θ≤ 180^(∘).
a Find the exact value of sin 2θ.
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b Find the exact value of cos 2θ.
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c Find the exact value of tan 2θ.
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Hint
a Use the Pythagorean Identity to find the value of cosθ.
b Recall the Double-Angle Identity for cosine.
c Apply the Tangent Identity.
Solution
a Start by recalling the Double-Angle Identity for sine.
sin^2 θ+cos^2 θ=1
Substitute
sinθ= 1/3
( 1/3)^2+cos^2 θ=1
PowQuot
(a/b)^m=a^m/b^m
1^2/3^2+cos^2 θ=1
CalcPow
Calculate power
1/9+cos^2 θ=1
SubEqn
LHS-1/9=RHS-1/9
cos^2 θ=1-1/9
OneToFrac
Rewrite 1 as 9/9
cos^2 θ=9/9-1/9
SubFrac
Subtract fractions
cos^2 θ=8/9
SqrtEqn
sqrt(LHS)=sqrt(RHS)
|cosθ|=sqrt(8/9)
▼
Calculate root
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
|cosθ|=sqrt(8)/sqrt(9)
SplitIntoFactors
Split into factors
|cosθ|=sqrt(4* 2)/sqrt(9)
CalcRoot
Calculate root
|cosθ|=2sqrt(2)/3
sin 2θ=2sinθcosθ
SubstituteII
sinθ= 1/3, cosθ= - 2sqrt(2)/3
sin 2θ=2( 1/3)( - 2sqrt(2)/3)
MultPosNeg
a(- b)=- a * b
sin 2θ=- 2(1/3)(2sqrt(2)/3)
MultFrac
Multiply fractions
sin 2θ=- 2(2sqrt(2)/9)
MoveLeftFacToNum
a*b/c= a* b/c
sin 2θ=- 4sqrt(2)/9
b In order to find the value of cos2θ, rewrite it by using the Double-Angle Identity for cosine.
cos 2θ=cos^2 θ-sin^2 θ
SubstituteII
sinθ= 1/3, cosθ= - 2sqrt(2)/3
cos 2θ=( - 2sqrt(2)/3)^2-( 1/3)^2
NegBaseToPosBase
(- a)^2=a^2
cos 2θ=(2sqrt(2)/3)^2-(1/3)^2
PowQuot
(a/b)^m=a^m/b^m
cos 2θ=(2sqrt(2))^2/3^2-1^2/3^2
CalcPow
Calculate power
cos 2θ=4* 2/9-1/9
Multiply
Multiply
cos 2θ=8/9-1/9
SubFrac
Subtract fractions
cos 2θ=7/9
c Finally, to find the value of tan 2θ, use the Tangent Identity.
tan 2θ = sin 2θ/cos 2θ
SubstituteII
sin 2θ= - 4sqrt(2)/9, cos 2θ= 7/9
tan 2θ = - 4sqrt(2)/9/7/9
DivFracByFracSL
.a/b /c/d.=a/b*d/c
tan 2θ = - 4sqrt(2)/9* 9/7
MultFrac
Multiply fractions
tan 2θ = - 4sqrt(2)* 9/9* 7
CrossCommonFac
Cross out common factors
tan 2θ = - 4sqrt(2)* 9/9* 7
CancelCommonFac
Cancel out common factors
tan 2θ = - 4sqrt(2)/7
Discussion
Presenting Half-Angle Identities
Before moving to the next station, the class stopped at another infopoint to learn about the Half-Angle Identities.
Rule
Half-Angle Identities
The half-angle identities are special cases of angle difference identities. To evaluate trigonometric functions of half an angle, the following identities can be applied.
sin θ/2 &= ±sqrt(1 - cosθ/2) [0.75em] cos θ/2 &= ±sqrt(1 + cosθ/2) [0.75em] tan θ/2 &= ±sqrt(1 - cosθ/1 + cosθ)
The sign of each formula is determined by the quadrant where the angle θ2 lies.
These identities are useful when finding the exact value of the sine, cosine, or tangent at a given angle.
Proof
Half-Angle IdentitiesFirst, write two of the Double-Angle Identities for cosine. cos 2x &= 1 - 2sin^2 x cos 2x &= 2cos^2 x - 1
Sine Identity
Start by solving the first identity written above for sin x.cos 2x = 1 - 2sin^2 x
▼
Solve for sin x
SubEqn
LHS-1=RHS-1
cos 2x - 1 = -2sin^2 x
DivEqn
.LHS /(-2).=.RHS /(-2).
cos 2x - 1/-2 = sin^2 x
RearrangeEqn
Rearrange equation
sin^2 x = cos 2x - 1/-2
MoveNegDenomToFrac
Put minus sign in front of fraction
sin^2 x = -cos 2x - 1/2
DistrNegSignSwap
-(b-a)=a-b
sin^2 x = 1-cos 2x/2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
sin x = ±sqrt(1-cos 2x/2)
sin θ/2 &= ±sqrt(1-cos(2* θ2)/2) & ⇓ sin θ/2 &= ±sqrt(1-cosθ/2) ✓
Cosine Identity
Start by solving the second identity written at the beginning for cos x.cos 2x = 2cos^2 x - 1
▼
Solve for cos x
AddEqn
LHS+1=RHS+1
1+cos 2x = 2cos^2 x
DivEqn
.LHS /2.=.RHS /2.
1+cos 2x/2 = cos^2 x
RearrangeEqn
Rearrange equation
cos^2 x = 1+cos 2x/2
SqrtEqn
sqrt(LHS)=sqrt(RHS)
cos x = ±sqrt(1+cos 2x/2)
cos θ/2 &= ±sqrt(1+cos (2* θ2)/2) & ⇓ cos θ/2 &= ±sqrt(1+cosθ/2) ✓
Tangent Identity
To derive the tangent identity, start by recalling the definition of the tangent ratio. tan x = sin x/cos x Next, substitute θ2 for x. tan θ2 = sin ( θ2)/cos ( θ2) Finally, substitute the half-identities for the sine and cosine into the equation above and simplify the right-hand side.tan θ/2 = sin ( θ2)/cos ( θ2)
SubstituteII
sin θ2= ±sqrt(1 - cosθ/2), cos θ/2= ±sqrt(1 + cosθ/2)
tan θ/2 = ±sqrt(1 - cosθ2)/±sqrt(1 + cosθ2)
▼
Simplify right-hand side
DivSqrt
sqrt(a)/sqrt(b)=sqrt(a/b)
tan θ/2 = ±sqrt(1-cosθ2/1+cosθ2)
DivFracByFracD
.a /b./.c /d.=a/b*d/c
tan θ/2 = ±sqrt(1-cosθ/2 * 2/1+cosθ)
CrossCommonFac
Cross out common factors
tan θ/2 = ±sqrt(1-cosθ/2 * 2/1+cosθ)
CancelCommonFac
Cancel out common factors
tan θ/2 = ±sqrt(1-cosθ/1 * 1/1+cosθ)
MultFrac
Multiply fractions
tan θ/2 = ±sqrt(1-cosθ/1+cosθ) ✓
Extra
Calculating cos 15^(∘)Consider the calculation of the exact value of cos 15^(∘).
- To be able to use the half-angle identities, the angle 15^(∘) needs to be rewritten as a certain angle divided by 2. Therefore, rewrite 15^(∘) as 30^(∘)2.
- Based on the trigonometric ratios of common angles, it is known that cos 30^(∘) = sqrt(3)2.
- According to the diagram of the quadrants, an angle that measures 15^(∘) is in the first quadrant. Therefore, the cosine ratio is positive.
cos 15^(∘)
Rewrite
Rewrite 15^(∘) as 30^(∘)/2
cos30^(∘)/2
cos θ/2=sqrt(1+cos θ/2)
sqrt(1+cos 30^(∘)/2)
cos 30^(∘) =sqrt(3)/2
sqrt(1+ sqrt(3)2/2)
sqrt(2+sqrt(3))/2
Example
Solving the Tasks Found Inside the Balloons
Zosia and her classmates reached the third site, ready for any task. Three balloons floated steadily in mist. They caught the balloons one by one and noticed that there is something inside each one. Zosia found a needle and popped the balloons.
Three notes fell to the ground. The classmates have been tasked with finding the values of the following trigonometric expressions by using the Half-Angle Identities. The answers must be written without any radicals in the denominators.
External credits: @brgfx
a sin 15^(∘)
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b cos 22.5^(∘)
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c tan (- 15^(∘))
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Solution
a First, review the Half-Angle Identity for sine.
sin(θ/2)=± sqrt(1-cosθ/2)
Substitute
θ= 30^(∘)
sin(30^(∘)/2)=± sqrt(1-cos 30^(∘)/2)
CalcQuot
Calculate quotient
sin 15^(∘)=± sqrt(1-cos 30^(∘)/2)
\ifnumequal{30}{0}{\cos\left(0^\circ\right)=1}{}\ifnumequal{30}{30}{\cos\left(30^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{45}{\cos\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\cos\left(60^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{90}{\cos\left(90^\circ\right)=0}{}\ifnumequal{30}{120}{\cos\left(120^\circ\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{30}{135}{\cos\left(135^\circ\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\cos\left(150^\circ\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{180}{\cos\left(180^\circ\right)=\text{-} 1}{}\ifnumequal{30}{210}{\cos\left(210^\circ\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{30}{225}{\cos\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\cos\left(240^\circ\right)=\text{-} \dfrac {1}2}{}\ifnumequal{30}{270}{\cos\left(270^\circ\right)=0}{}\ifnumequal{30}{300}{\cos\left(300^\circ\right)=\dfrac{1}2}{}\ifnumequal{30}{315}{\cos\left(315^\circ\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\cos\left(330^\circ\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{30}{360}{\cos\left(360^\circ\right)=1}{}
sin 15^(∘)=± sqrt(1-sqrt(3)/2/2)
1=a/a
sin 15^(∘)=± sqrt(2/2-sqrt(3)/2/2)
SubFrac
Subtract fractions
sin 15^(∘)=± sqrt(2-sqrt(3)/2/2)
DivFracSL
.a/b /c.= a/b* c
sin 15^(∘)=± sqrt(2-sqrt(3)/4)
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
sin 15^(∘)=± sqrt(2-sqrt(3))/sqrt(4)
CalcRoot
Calculate root
sin 15^(∘)=± sqrt(2-sqrt(3))/2
b To calculate the cosine of 22.5^(∘), use the Half-Angle Identity for cosine.
cos(θ/2)=± sqrt(1+cosθ/2)
Substitute
θ= 45^(∘)
cos(45^(∘)/2)=± sqrt(1+cos 45^(∘)/2)
CalcQuot
Calculate quotient
cos 22.5^(∘)=± sqrt(1+cos 45^(∘)/2)
\ifnumequal{45}{0}{\cos\left(0^\circ\right)=1}{}\ifnumequal{45}{30}{\cos\left(30^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{45}{45}{\cos\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{45}{60}{\cos\left(60^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{45}{90}{\cos\left(90^\circ\right)=0}{}\ifnumequal{45}{120}{\cos\left(120^\circ\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{45}{135}{\cos\left(135^\circ\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{45}{150}{\cos\left(150^\circ\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{45}{180}{\cos\left(180^\circ\right)=\text{-} 1}{}\ifnumequal{45}{210}{\cos\left(210^\circ\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{45}{225}{\cos\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{45}{240}{\cos\left(240^\circ\right)=\text{-} \dfrac {1}2}{}\ifnumequal{45}{270}{\cos\left(270^\circ\right)=0}{}\ifnumequal{45}{300}{\cos\left(300^\circ\right)=\dfrac{1}2}{}\ifnumequal{45}{315}{\cos\left(315^\circ\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{45}{330}{\cos\left(330^\circ\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{45}{360}{\cos\left(360^\circ\right)=1}{}
cos 22.5^(∘)=± sqrt(1+sqrt(2)/2/2)
1=a/a
cos 22.5^(∘)=± sqrt(2/2+sqrt(2)/2/2)
AddFrac
Add fractions
cos 22.5^(∘)=± sqrt(2+sqrt(2)/2/2)
DivFracSL
.a/b /c.= a/b* c
cos 22.5^(∘)=± sqrt(2+sqrt(2)/4)
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
cos 22.5^(∘)=± sqrt(2+sqrt(2))/sqrt(4)
CalcRoot
Calculate root
cos 22.5^(∘)=± sqrt(2+sqrt(2))/2
c The value of tan(- 15^(∘)) can be calculated by applying the Half-Angle Identity for tangent.
tan(- 30/2)=±sqrt(1-cos 30^(∘)/1+cos 30^(∘))
CalcQuot
Calculate quotient
tan(- 15^(∘))=±sqrt(1-cos 30^(∘)/1+cos 30^(∘))
\ifnumequal{30}{0}{\cos\left(0^\circ\right)=1}{}\ifnumequal{30}{30}{\cos\left(30^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{45}{\cos\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\cos\left(60^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{90}{\cos\left(90^\circ\right)=0}{}\ifnumequal{30}{120}{\cos\left(120^\circ\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{30}{135}{\cos\left(135^\circ\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\cos\left(150^\circ\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{180}{\cos\left(180^\circ\right)=\text{-} 1}{}\ifnumequal{30}{210}{\cos\left(210^\circ\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{30}{225}{\cos\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\cos\left(240^\circ\right)=\text{-} \dfrac {1}2}{}\ifnumequal{30}{270}{\cos\left(270^\circ\right)=0}{}\ifnumequal{30}{300}{\cos\left(300^\circ\right)=\dfrac{1}2}{}\ifnumequal{30}{315}{\cos\left(315^\circ\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\cos\left(330^\circ\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{30}{360}{\cos\left(360^\circ\right)=1}{}
tan(- 15^(∘))=±sqrt(1-sqrt(3)/2/1+sqrt(3)/2)
1=a/a
tan(- 15^(∘))=±sqrt(2/2-sqrt(3)/2/2/2+sqrt(3)/2)
AddSubFrac
Add and subtract fractions
tan(- 15^(∘))=±sqrt(2-sqrt(3)/2/2+sqrt(3)/2)
DivFracByFracSL
.a/b /c/d.=a/b*d/c
tan(- 15^(∘))=±sqrt(2-sqrt(3)/2* 2/2+sqrt(3))
MultFrac
Multiply fractions
tan(- 15^(∘))=±sqrt((2-sqrt(3))2/2(2+sqrt(3)))
ReduceFrac
a/b=.a /2./.b /2.
tan(- 15^(∘))=±sqrt(2-sqrt(3)/2+sqrt(3))
tan(- 15^(∘))=±sqrt(2-sqrt(3)/2+sqrt(3))
ExpandFrac
a/b=a * (2-sqrt(3))/b * (2-sqrt(3))
tan(- 15^(∘))=±sqrt((2-sqrt(3)) (2-sqrt(3))/(2+sqrt(3)) (2-sqrt(3)))
▼
Simplify right-hand side
ExpandDiffSquares
(a+b)(a-b)=a^2-b^2
tan(- 15^(∘))=±sqrt((2-sqrt(3))(2-sqrt(3))/2^2-(sqrt(3))^2)
ProdToPowTwoFac
a* a=a^2
tan(- 15^(∘))=±sqrt((2-sqrt(3))^2/2^2-(sqrt(3))^2)
ExpandNegPerfectSquare
(a-b)^2=a^2-2ab+b^2
tan(- 15^(∘))=±sqrt(2^2-4sqrt(3)+(sqrt(3))^2/2^2-(sqrt(3))^2)
CalcPow
Calculate power
tan(- 15^(∘))=±sqrt(4-4sqrt(3)+3/4-3)
AddSubTerms
Add and subtract terms
tan(- 15^(∘))=±sqrt(7-4sqrt(3)/1)
DivByOne
a/1=a
tan(- 15^(∘))=±sqrt(7-4sqrt(3))
Example
Solving a Task from a Fortune Cookie
The class has successfully made it to the fourth mathematical sacred site which appears to be in a kitchen. Something seems strange. They door immediately close behind them. They are not in a kitchen at all, but an escape room! In order to get out, the class needed to find the task and solve it.
After searching for a while, someone found a fortune cookie. Inside of it, the task challenges them to find the exact value of two expressions given that sinθ=- 1517 and 270^(∘)≤θ≤360^(∘).
a sin θ/2
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b cos θ/2
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Hint
a Start by calculating cosθ by using the Pythagorean Identity.
b Use the Half-Angle Identity for cosine.
Solution
a In order to find the value of sin θ2, the Half-Angle Identity for sine can be used.
sin^2 θ+cos^2 θ=1
Substitute
sinθ= - 15/17
( - 15/17)^2+cos^2 θ=1
▼
Solve for cosθ
NegBaseToPosPow
(- a)^2 = a^2
(15/17)^2+cos^2 θ=1
PowQuot
(a/b)^m=a^m/b^m
15^2/17^2+cos^2 θ=1
CalcPow
Calculate power
225/289+cos^2 θ=1
SubEqn
LHS-225/289=RHS-225/289
cos^2 θ=1-225/289
OneToFrac
Rewrite 1 as 289/289
cos^2 θ=289/289-225/289
SubFrac
Subtract fractions
cos^2 θ=64/289
SqrtEqn
sqrt(LHS)=sqrt(RHS)
|cos θ|=sqrt(64/289)
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
|cos θ|=sqrt(64)/sqrt(289)
CalcRoot
Calculate root
|cos θ|=8/17
sin θ/2=±sqrt(1-cosθ/2)
Substitute
cosθ= 8/17
sin θ/2=±sqrt(1- 8/17/2)
OneToFrac
Rewrite 1 as 17/17
sin θ/2=±sqrt(17/17-8/17/2)
SubFrac
Subtract fractions
sin θ/2=±sqrt(9/17/2)
DivFracSL
.a/b /c.= a/b* c
sin θ/2=±sqrt(9/34)
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
sin θ/2=±sqrt(9)/sqrt(34)
CalcRoot
Calculate root
sin θ/2=±3/sqrt(34)
b To find the value of cos θ2, recall the Half-Angle Identity for cosine.
cos θ/2=±sqrt(1+cosθ/2)
Substitute
cosθ= 8/17
cos θ/2=±sqrt(1+ 8/17/2)
OneToFrac
Rewrite 1 as 17/17
cos θ/2=±sqrt(17/17+8/17/2)
AddFrac
Add fractions
cos θ/2=±sqrt(25/17/2)
DivFracSL
.a/b /c.= a/b* c
cos θ/2=±sqrt(25/34)
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
cos θ/2=±sqrt(25)/sqrt(34)
CalcRoot
Calculate root
cos θ/2=±5/sqrt(34)
Example
Choosing the Right Direction
The class successfully solved the task from the fortune cookie and arrived at the fifth site. There they saw two potential roads to the next site. To determine which road to choose, they need to solve the task written on the placard.
Which way should the class turn to end up at the sixth site?
External credits: @brgfx
{"type":"choice","form":{"alts":["Left","Right"],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":1}
Hint
Use the Pythagorean Identity and the Double-Angle Identity for sine.
Solution
To verify the identity, rewrite its sides until they match. First, add sin^2 2x and subtract 4cos^4 x from both sides of the equation. Then, factor out 4cos^2 x.
Recall the Pythagorean Identity and rewrite it to match the expression in the parentheses.
sin^2 x+cos^2 x=1 ⇓ sin^2 x=1-cos^2 x
Now, substitute 1-cos^2 x for sin^2 x into the equation and simplify it. Next, use the Double-Angle Identity for sine.
As can be seen, the left-hand side is the same as the right-hand side. Therefore, the identity is verified. This means that the class should turn right to end up at the sixth site.
4cos^2 x-sin^2 2x=4cos^4 x
AddEqn
LHS+sin^2 2x=RHS+sin^2 2x
4cos^2 x=4cos^4 x+sin^2 2x
SubEqn
LHS-4cos^4 x=RHS-4cos^4 x
4cos^2 x-4cos^4 x=sin^2 2x
FactorOut
Factor out 4cos^2 x
4cos^2 x(1-cos^2 x)=sin^2 2x
4cos^2 x(1-cos^2 x)=sin^2 2x
1 - cos^2(θ) = sin^2(θ)
4cos^2 xsin^2 x=sin^2 2x
sin^2(θ)=(sin(θ))^2
4cos^2 xsin^2 x=(sin 2x)^2
sin(2θ)=2sin(θ)cos(θ)
4cos^2 xsin^2 x=(2sin xcos x)^2
PowProd
(a * b)^m=a^m* b^m
4cos^2 xsin^2 x=4sin^2 xcos^2 x
Example
Matching the Sides of Identities
The class made the correct turn to the right and arrived at the sixth site. They notice a rustic table with torn pieces of paper, on which the left-hand and right-hand sides of different identities were written. The task at hand is to match at least one of the identities.
Zosia placed two random pieces in the middle of the table and started checking whether they form an identity. sinθ/2cosθ/2? =sinθ/2
What result is she going to get?
{"type":"choice","form":{"alts":["It is an identitiy.","It is not an identity."],"noSort":false},"formTextBefore":"","formTextAfter":"","answer":0}
Hint
Rewrite one or both sides of the equation by using the Half-Angle Identity or the Double-Angle Identity.
Solution
In order to check whether the equation Zosia formed is an identity, rewrite the sides until they match. There are two ways to do this — using the Half-Angle Identity or using the Double-Angle Identity. Each way will be shown one at a time.
Using the Half-Angle Identity
Start by recalling the Half-Angle Identities for sine and cosine. sin θ/2=± sqrt(1-cosθ/2) [0.2cm] cos θ/2=± sqrt(1+cosθ/2) Substitute the expressions corresponding to sin θ2 and cos θ2 into the equation and simplify.sinθ/2cosθ/2? =sinθ/2
SubstituteExpressions
Substitute expressions
± sqrt(1-cosθ/2)(± sqrt(1+cosθ/2))? =sinθ/2
ProdSqrt
sqrt(a)*sqrt(b)=sqrt(a* b)
± sqrt(1-cosθ/2*1+cosθ/2)? =sinθ/2
MultFrac
Multiply fractions
± sqrt((1-cosθ)(1+cosθ)/4)? =sinθ/2
(a-b)(a+b)=a^2-b^2
± sqrt(1^2-cos^2 θ/4)? =sinθ/2
BaseOne
1^a=1
± sqrt(1-cos^2 θ/4)? =sinθ/2
± sqrt(1-cos^2 θ/4)? =sinθ/2
1 - cos^2(θ) = sin^2(θ)
± sqrt(sin^2 θ/4)? =sinθ/2
SqrtQuot
sqrt(a/b)=sqrt(a)/sqrt(b)
± sqrt(sin^2 θ)/sqrt(4)? =sinθ/2
CalcRoot
Calculate root
± sinθ/2=sinθ/2 ✓
Using the Double-Angle Identity
Start by reviewing the Double-Angle Identity. sin 2θ=2sinθcosθ If instead of 2θ there was θ, the identity would look the following way. sin θ=2sinθ/2cos θ/2 Finally, by dividing both sides of the identity by 2, the equation that should have been verified can be obtained. sinθ/2=sinθ/2cos θ/2 Therefore, the equation Zosia formed is indeed an identity.
Example
Simplifying Trigonometric Expressions
The class has arrived at the seventh mathematical sacred sit — the finale. A huge board with plenty of colorful stickers lies ahead. They are told to choose two random cards.
The classmates go through and and turn the cards. Lo and behold, two trigonometric expressions appear that need to be simplified. Once this is done, they will have completed the math quest!
External credits: @brgfx
a sin^2 θ/2-cos^2 θ/2
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b cos2θ/sinθ+cosθ
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Hint
a Start by factoring out - 1.
b Use the Double-Angle Identity.
Solution
a Start by analyzing the first given expression.
b Again, to simplify the expression, use the Double-Angle Identity for cosine.
cos2θ/sinθ+cosθ
cos(2θ)=cos^2(θ)-sin^2(θ)
cos^2 θ-sin^2 θ/sinθ+cosθ
FacDiffSquares
a^2-b^2=(a+b)(a-b)
(cosθ+sinθ)(cosθ-sinθ)/sinθ+cosθ
AssociativePropAdd
Associative Property of Addition
(cosθ+sinθ)(cosθ-sinθ)/cosθ+sinθ
CrossCommonFac
Cross out common factors
(cosθ+sinθ)(cosθ-sinθ)/(cosθ+sinθ)
CancelCommonFac
Cancel out common factors
cosθ-sinθ/1
DivByOne
a/1=a
cosθ-sinθ
Closure
Solving the Task to Get the Bonus
When the math quest game for Zosia and her classmates began, they were going to face some tough tasks. After passing all the sites, the rules stated that they could try for a bonus task and get major points. Naturally, they decided try to solve it.
What is the value of HD?
External credits: @brgfx
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Hint
Use the Double-Angle Identity for sine and the Tangent Identity.
Solution
To find the value of HD, substitute the corresponding expressions to H and D and then simplify.
Now, the Double-Angle Identity for sine can be used.
Therefore, the quotient of H and D simplifies to 14tan θ.
H/D
SubstituteII
H= v^2/2gsin^2 θ, D= v^2/gsin2θ
v^2/2gsin^2 θ/v^2/gsin2θ
MoveRightFacToNum
a/c* b = a* b/c
v^2sin^2 θ/2g/v^2sin 2θ/g
DivFracByFracD
.a /b./.c /d.=a/b*d/c
v^2sin^2 θ/2g* g/v^2sin2θ
MultFrac
Multiply fractions
v^2sin^2 θ g/2gv^2sin2θ
CrossCommonFac
Cross out common factors
v^2sin^2 θ g/2gv^2sin2θ
CancelCommonFac
Cancel out common factors
sin^2 θ/2sin2θ
sin^2 θ/2sin2θ
sin(2θ)=2sin(θ)cos(θ)
sin^2 θ/2* 2sinθcosθ
Multiply
Multiply
sin^2 θ/4sinθcosθ
CrossCommonFac
Cross out common factors
\dfrac{\sin^\cancel{2}\!\theta }{4\cancel{\sin\theta}\cos\theta}
CancelCommonFac
Cancel out common factors
sinθ/4cosθ
WriteProdFrac
Write as a product of fractions
1/4*sinθ/cosθ
tan(θ)=sin(θ)/cos(θ)
1/4tan θ
Double and Half-Angle Identities
Exercise 3.1
If cos 2a=cos 2b, does it also always mean that a=b?
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Let's assume that cos2a is equal to cos2b. We will discuss if this statement always implies that a=b. Let's first use the Double Angle Identity for cosine to express cos2a and cos2b in terms of the trigonometric ratios of the angles a and b. cos2 a &= cos^2 a-sin^2 a cos2 b &= cos^2 b-sin^2 b Because of how cosine values on the unit circle relate to one another, let's arbitrarily take two supplementary angles, a= 30^(∘) and b= 150^(∘), and substitute them into our equations. cos2( 30^(∘)) &= cos^2 30^(∘)-sin^2 30^(∘) cos2( 150^(∘)) &= cos^2 150^(∘)-sin^2 150^(∘) Now, we will analyze the angles of 30^(∘) and 150^(∘) on a unit circle. Let A and B be the points that these angles create on the unit circle, respectively.
Then, the coordinates of A and B can be expressed as follows. A(cos 30^(∘), sin 30^(∘)) B(cos 150^(∘), sin 150^(∘)) From the diagram, we can see that A and B have the same y-coordinate and opposite x-coordinates.
This means that the following statements are true. cos 30^(∘) &=- cos 150^(∘) sin 30^(∘) &= sin 150^(∘) With this information, we can substitute cos 30^(∘) with - cos 150^(∘) and sin 30^(∘) with sin 150^(∘).
Note that the right-hand side of this equation is the same as the right-hand side of the equation for cos 2(150^(∘)). cos2(30^(∘)) &= cos^2 150^(∘)-sin^2 150^(∘) cos2(150^(∘)) &= cos^2 150^(∘)-sin^2 150^(∘) This means that the left-hand sides are also equal. cos2( 30^(∘))&=cos2( 150^(∘)) &⇕ cos 60^(∘) &= cos 300^(∘) However, a= 30^(∘) is not equal to b= 150^(∘). Therefore, we can conclude that even though cos 2a=cos 2b, a does not have to be equal to b.
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Double and Half-Angle Identities
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