Double Angle and Half Angle Formulas

Did you know that

Get started Sign up for free
Double Angle and Half Angle Formulas Double Angle and Half Angle Formulas

Create learning materials about Double Angle and Half Angle Formulas with our free learning app!

  • Instand access to millions of learning materials
  • Flashcards, notes, mock-exams and more
  • Everything you need to ace your exams
Create a free account

Millions of flashcards designed to help you ace your studies

Sign up for free

Convert documents into flashcards for free with AI!

Contents
Table of contents

    sin2θ2sinθ

    and

    cosθ2cosθ2?

    In this article, you would understand what happens when trigonometric identities are either doubled or halved.

    Double-angle formulas

    Trigonometric functions can be doubled but not in the same way as normal numbers are doubled.

    If you have the expression 3y and you are to double it, it is easy to multiply 3y by 2 to get 6y. Note that sin30° is 0.5 and doubling the angle gives 60°, but sin60° would not give you 1. As normal mathematical operations would multiply 0.5 by 2 to give 1, trigonometric identities would require their own formula to double their function.

    Deriving the double-angle formula for Sine function

    We intend on finding the formula for sin2θ. Note that

    sin2θ=sin(θ+θ)

    Recall that,

    sin(A+B)=sinAcosB+sinBcosA

    Now takeA=B=θ, we get

    sin2θ=sinθcosθ+sinθcosθ =2sinθcosθ

    The double angle formula for the sine function is given by sin (2θ)=2 sinθ cosθ.

    Findsin 60 °using the double angle formula.

    Solution:

    We have

    60°=2(30°)

    Thus

    sin 60°=sin (2×30°) =2 sin 30°cos 30°

    but,

    sin 30°=12, cos 30°=32

    Then,

    sin 60°=2×12×32sin 60°=32

    Given that90°<θ<180°, find sin 2θ if

    sinθ=45

    Solution:

    We have sinθ in the given, but in order to apply our formula, we need to find cosθ.

    Recall that,

    cos2θ+sin2θ=1

    Thus,

    cos2θ=1-sin2θ=1-(45)2 =1-1625cos2θ=925

    We take the square root of both sides to get,

    cosθ=±35

    Note that the range of the angle is between 90° and 180°, this means that θ is in the second quadrant. The cosine of angles in the second quadrant has negative values. Thus,

    cosθ=-35

    Now we have to apply our double angle formula,

    sin2θ=2sinθcosθ =2×45×(-35) =-2425

    Deriving the double-angle formula for Cosine function

    We will now develop a formula for a double-angle for the cosine function. We derive three equal formulas.

    We first note that

    cos2θ=cos(θ+θ)

    Now, recall that

    cos(A+B)=cosAcosB-sinAsinB

    By takingA=B=θ,we get

    cos2θ= cos(θ+θ) = cosθcosθ-sinθsinθ = cos2θ-sin2θ

    Thus, we derive the first formula for cos 2θ,

    cos 2θ=cos2θ-sin2θ

    We now recall the identity cos2θ+sin2θ=1, thus we havecos2θ=1-sin2θ .

    Now we replace this with the obtained formula forcos 2θ, to get

    cos2θ=1-sin2θ-sin2θ =1-2sin2θ

    Thus, the second formula for cos 2θ is

    cos 2θ=1-2sin2θ

    In a similar way, we have sin2θ=1-cos2θ.

    Substituting the value of sin2θ into the formula of cos 2θ, we have

    cos2θ=cos2θ-(1-cos2θ) =cos2θ-1+cos2θ =2cos2θ-1

    Thus, the third formula for cos 2θ is

    cos 2θ=2 cos2θ-1

    The double angle formulas for the cosine function are given by,

    cos 2θ=cos2θ-sin2θ =2 cos2θ-1 =1-2sin2θ

    Given that 90°<θ< 180°, find cos 2θ if

    sinθ=45

    Solution:

    Method 1.

    The direct way to find cos 2θ is to use the formula cos 2θ=1-2sin2θ, since we are given the value of sinθ.

    So,

    cos 2θ=1-2sin2θ =1-2452 =1-21625 =1-3225 = 25-3225 =-725

    Method 2.

    We can use either of the other formulas to find id="2967226" role="math" cos 2θ, we will use id="2967228" role="math" cos 2θ=2cos2θ-1.We need thus to find cos θ.

    We recall that cos2θ+sin2θ=1, thus

    cos2θ=1-sin2θ =1-452 =1-1625 =925

    Taking the square root of both sides, we get

    cosθ=±35

    Note that 90°<θ<180°, this means that θ is in the second quadrant. The cosine of angles in the second quadrant have negative values. Thus,

    cosθ=-35

    So, we can apply our formula

    cos2θ=2cos2θ-1 = 2-352-1 = 2×925-1 =1825-1 =-725

    For 180°<θ<270°, find cos 2θ when

    cosθ=-13

    Solution:

    In solving this problem, it is faster to use the formulacos2θ=2cos2θ-1.

    Thus,

    cos2θ=2×-132-1 =2×19-1 =29-1 =-79

    Deriving the double-angle formula for Tangent function

    We will develop a formula for a double-angle of the tangent function.

    We recall that

    tanθ=sinθcosθ

    and,

    sin2θ=2sinθcosθ

    cos2θ=cos2θ-sin2θ

    Thus,

    tan2θ=sin2θcos2θ

    Substituting sin 2θ and cos 2θ by their expressions, we get

    tan2θ=2sinθcosθcos2θ-sin2θ

    To simplify this further, we divide both the numerator and the denominator of the right-hand side of the equation by cos2θ, to get

    tan2θ=2sinθcosθcos2θcos2θ-sin2θcos2θ =2sinθcosθcos2θcos2θ-sin2θcos2θ =2tanθ1-tan2θ

    The double angle formula for the tangent function is given by,

    tan 2θ=2 tan θ1-tan2θ

    Given that 90°<θ<180°, find tan 2θ if

    sin θ=45

    Solution:

    We need to findtan θ, this means that we firstly have to find cos θ.

    We recall that cos2θ+sin2θ=1, thus cos2θ=1-sin2θ. Replacing sin θ by its value, we get

    cos2θ=1-452 =1-1625 =925

    We take the square root of both sides, to get

    cosθ=±35

    Note that 90°<θ<180°, this means that θ is in the second quadrant. The cosine of angles in the second quadrant have negative values. Thus, cosθ=-35.

    Therefore,

    tanθ=sinθcosθ =45-35 =45×(-53) -43

    Thus,

    tan2θ=2tanθ1-tan2θ =2×-431--432 =-831-169 =-83-79 =-83×-97 =247

    Deriving the double-angle formulas for Secant, Cosecant and Cotangent functions

    Secant, cosecant, cotangent functions are the reciprocals of cosine, sine and tangent respectively. In order to derive their double-angle formulas, you just need to find the multiplicative inverse of the corresponding double-angle formulas.

    Double angle formula for Secant

    We recall by the definition of the secant function that

    secθ=1cosθ so sec 2θ=1cos 2θ

    but from the double angle formula for cosine we have cos 2θ=cos2θ-sin2θ , thus

    sec 2θ=1cos2θ-sin2θ

    Now let us express sec 2θin terms of sec θ and csc θ.

    In fact, cos θ=1sec and csc θ=1sin θ, thus we have

    sec 2θ=11sec θ2-1csc θ2 =11sec2θ-1csc2θ =1csc2θ-sec2θsec2θ csc2θ =sec2θ csc2θcsc2θ-sec2θ.

    The double angle formula for the secant function is given by,

    sec 2θ=sec2θ csc2θcsc2θ-sec2θ

    Double angle formula for Cosecant

    We recall by the definition of the secant function that

    csc θ=1sinθso csc 2θ=1sin 2θ

    but from the double angle formula for sine function, we havesin 2θ=2 sin θ cos θ, thus

    csc 2θ=12sin θcos θ =12×1sin θ×1cos θ =12×csc θ ×sec θcsc 2θ=12csc θ sec θ

    The double angle formula for the cosecant function is given by,

    csc 2θ=12csc θ sec θ

    Double angle formula for Cotangent

    We recall by the definition of the secant function that

    cot θ=1tan θso cot 2θ=1tan 2θ

    We recall from the double angle formula for the tangent function that tan 2θ=2 tan θ1-tan2θ, we have

    cot 2θ=12 tan θ1-tan2θ =1-tan2θ2 tan θ

    The double angle formula for the cotangent function is given by,

    cot 2θ=1-tan2θ2 tan θ

    Given that 90°<θ<180°, find sec 2θ, csc 2θ and cot 2θ given that

    sinθ=45

    Solution:

    We have the value of sinθ, but in order to apply these formulas, we need to find cosθ.

    We recall that,cos2θ+sin2θ=1

    cos2θ=1-sin2θ=1-452=1-1625=925

    Thuscosθ=±35, but since90°<θ<180°,thuscosθ=-35.

    So

    sec θ=-53, csc θ=54

    Hence, we have

    sec 2θ=sec2θ csc2θcsc2θ-sec2θ=-532542542--532=259×25162516-259=625144225-400144=625-175=-257

    csc 2θ=1sin 2θ =12cosθsinθ=12-3545=1-2425=-2524cot2θ=1-tan2θ2tanθ, but tan θ=sin θcos θ=45-35=-43, thus we have cot2θ=1--4322-43=1-169-83=-79-83=-79×(-38)=724.

    Half-angle formulas

    Trigonometric functions can be halved but not in the same manner normal numbers are halved. If you have the expression 6y and you are to half it, it is easy to multiply 6y by 0.5 to get 3y. Note that sin30° is 0.5 and halving the angle gives 15 degrees, but sin15° would not give you 0.25. As normal mathematical operations would multiply 0.5 by 0.5 (half) to give 0.25, trigonometric identities would require their own formula to half its function.

    Deriving the half-angle formula for Sine

    To find sinθ2, we recall first that

    cos2θ=1-2sin2θ

    Let θ=2, thus

    cos2×ϕ2=1-2sin2ϕ2 cosϕ=1-2sin2ϕ2

    In order to isolate sin2ϕ2, subtract 1 from both sides, to get

    coϕ-1=1-2sin2ϕ2-1-2sin2ϕ2=cos ϕ-1

    We divide both sides of the equation by -2, we get

    sin2ϕ2=1-cosϕ2

    Taking the square root of both sides of the equation, we get

    sinϕ2=±1-cosϕ2

    The half-angle formula for the sine function is given by,

    sinϕ2=±1-cosϕ2

    If sinθ=23, and 90°<θ<180°, find sinθ2.

    Solution:

    Since 90°<θ<180°, 45°<θ2<90°, thus sin θ>0.Hence,

    sinθ2=1-cosθ2

    Thus, to findsinθ2, we need to find cosθ. Recall that

    cos2θ=1-sin2θcosθ=1-sin2θ

    Since

    sinθ=23

    Then,

    cosθ=1-(23)2cosθ=1-49cosθ=59=53

    Now we can substitute the value of cosθ into our equation

    sinθ2=1-cosθ2sinθ2=1-532sinθ2=3-532sinθ2=3-53×12sinθ2=3-56

    Deriving the half-angle formula for cosine

    Recall that

    cos2θ=2cos2θ-1

    Where,

    θ=2

    Therefore

    cos(2×2)=(2×cos22)-1cos=(2×cos22)-1

    Add 1 to both sides of the equation

    cos+1=(2×cos22)-1+1cos+1=2×cos22

    Divide both sides by 2

    cos+12=cos22

    Find the square root of both sides of the equation

    cos+12=cos2

    Thus

    cosθ2=±cosθ+12

    Given that

    sinθ=-34

    for 180°<θ<270°, find cosθ2.

    Solution:

    To begin, get the value of cosθ.

    Note that

    cos2θ=1-sin2θ

    Thus

    cos2θ=1-(-34)2 cos2θ=1-916cos2θ=716cosθ=±74

    Recall from the question that θ falls within the third quadrant, hence cosine values would be negative. Thus

    cosθ=-74

    Note before

    cosθ2=±cosθ+12

    So by substituting the value of cosθ we get

    cosθ2=±1-742cosθ2=±4-742cosθ2=±4-74×12cosθ2=±4-78cosθ2=±4-722

    Multiply the right hand side of the equation by 22 (rationalization of surds)

    cosθ2=±4-722×22cosθ2=±8-274

    Now θ has been halved, the conditions would change too by

    180°<θ<270°

    Angles here fall in the third quadrant.

    Dividing that by 2 you have

    90°<θ2<135°

    θ2 falls in the second quadrant and cosθ is negative in the second quadrant.

    cosθ2=-8-274

    Deriving the half-angle formula for tangents

    Knowing that

    tanθ=sinθcosθsinθ2=±1-cosθ2cosθ2=±cosθ+12

    Then

    tanθ2=1-cosθ2cosθ+12tanθ2=1-cosθ2cosθ+12tanθ2=1-cosθ2×2cosθ+1tanθ2=1-cosθcosθ+1

    Multiply the right-hand side of the equation by cosθ+1cosθ+1 and you would have

    tanθ2=1-cos2θcosθ+1

    Recall that

    1-cos2θ=sin2θ

    Then

    tanθ2=sin2θcosθ+1tanθ2=sinθcosθ+1

    Find tan2 when tan=43.

    Solution:

    With the value given, the opposite and adjacent are 4 and 3 respectively. Using the Pythagoras theorem we shall arrive at a value for the hypotenuse.

    hypotenuse2=opposite2+adjacent2hypotenuse2=42+32hypotenuse2=25hypotenuse=5

    Now we have the value of the hypotenuse, then

    sin=45cos=35

    You can now apply the formula

    tanθ2=sinθcosθ+1θ=tan2=sincos+1tan2=4535+1tan2=4585tan2=45×58tan2=12

    Deriving the half-angle formula for secant, cosecant and cotangent

    As mentioned earlier, secant, cosecant, cotangent are the inverse of cosine, sine and tangent respectively. So as to derive their half-angle formulas, you just need to find the multiplicative inverse of the corresponding half-angle formulas. Thus the half-angle formula of secant becomes:

    secθ=1cosθcosθ2=±cosθ+12secθ2=±2cosθ+1

    the half-angle formula of cosecant becomes:

    cosecθ=1sinθsinθ2=±1-cosθ2cosecθ2=±21-cosθ

    and the half-angle formula of cotangent becomes:

    cotθ=1tanθtanθ2=sinθcosθ+1cotθ2=cosθ+1sinθ

    This is the same as

    cotθ2=cosθ+1sinθcotθ2=cosθsinθ+1sinθcotθ2=cotθ+cosecθ

    If secθ=1312, find values for secθ2, cosecθ2 and cotθ2.

    Solution:

    Since,

    secθ=1312

    and

    secθ=1cosθ

    Then,

    cosθ=1213

    Knowing that;

    sin2θ=1-cos2θsinθ=1-cos2θ

    Therefore;

    sinθ=1-(1213)2sinθ=1-144169sinθ=25169sinθ=513

    Since the values for cosθ as well as sinθ have been found, it is easier to find the half angles of sec, cosec and cot. Thus half-angle of sec becomes:

    secθ2=±2cosθ+1secθ2=21213+1secθ2=22513secθ2=2×1325secθ2=2625secθ2=265

    For half-angle of cosec

    cosecθ2=±21-cosθcosecθ2=21-1213cosecθ2=2113cosecθ2=2×131cosecθ2=26

    And for the half-angle of cot

    cotθ2=cosθ+1sinθcotθ2=1213+1513cotθ2=2513513cotθ2=2513×135cotθ2=5

    Applications of double-angle and half-angle formulas

    Here are a few examples that show the application of double-angle and half-angle formulas.

    Solve the for θ in

    sin(2θ)+4sinθ+2cosθ=-4

    Solution:

    Recall that

    sin2θ=2sinθcosθ

    Substitute into the equation. Therefore,

    2sinθcosθ+4sinθ+2cosθ=-42sinθcosθ+4sinθ+2cosθ+4=0(2sinθcosθ+4sinθ)+(2cosθ+4)=02sinθ(cosθ+2)+2(cosθ+2)=0(2sinθ+2)(cosθ+2)=0

    This means that

    2sinθ+2=02sinθ=-22sinθ2=-22sinθ=-1

    or

    cosθ+2=0cosθ=-2

    Now in finding θ, we have to find both arcsinθ and arccosθ. Therefore,

    θ=sin-1-1θ=270°

    However, -2 goes beyond the possible values for arccosθ. Hence,

    cosθ=-2

    is invalid

    Thus the value of θ is 270°.

    If

    sinϕ=15

    find

    cosϕ2

    Solution:

    The first thing to do is to find cosϕ. Knowing that

    cos2ϕ+sin2ϕ=1cos2ϕ=1-sin2ϕcosϕ=1-sin2ϕ

    Now, substitute the value of sinϕ to find cosϕ.

    cosϕ=1-sin2ϕcosϕ=1-15cosϕ=45cosϕ=25cosϕ=25×55cosϕ=255

    Recall that

    cosϕ2=±cosθ+12

    Hence,

    cosϕ2=±255+12cosϕ2=±25+552cosϕ2=±25+55×12cosϕ2=±25+510

    Double-Angle and Half-Angle Formulas - Key takeaways

    • A trigonometric function cannot be halved or doubled using normal arithmetic methods. Rather, some formulas are needed to carry out such operations.
    • To double the angle of sine functions:sin2θ=2sinθcosθ
    • To double the angle of cosine functions, any of the following formulas can be used.cos2θ=cos2θ-sin2θ orcos2θ=1-2sin2θ or cos2θ=2cos2θ-1.
    • To double the angle of tangent functions: tan2θ=2tanθ1-tan2θ.
    • To double the angle of secant functions: sec2θ=1cos2θ-sin2θ.
    • To double the angle of cosecant functions: cosec2θ=12sinθcosθ.
    • To double the angle of cotangent functions: cot2θ=1-tan2θ2tanθ.
    • To find the half-angle of sine functions use: sinθ2=±1-cosθ2.
    • To find the half-angle of cosine function use: cosθ2=±cosθ+12
    • To find the half-angle of tangent functions use: tanθ2=sinθcosθ+1.
    • To find the half-angle of secant functions use:secθ2=±2cosθ+1.
    • To find the half-angle of cosecant functions use:cosecθ2=±21-cosθ.
    • To find the half-angle of cotangent functions use:cotθ2=cosθ+1sinθ.
    Double Angle and Half Angle Formulas Double Angle and Half Angle Formulas
    Learn with 0 Double Angle and Half Angle Formulas flashcards in the free StudySmarter app

    We have 14,000 flashcards about Dynamic Landscapes.

    Sign up with Email

    Already have an account? Log in

    Frequently Asked Questions about Double Angle and Half Angle Formulas

    What are double-angle and half-angle formulas?

    Double-angle and half-angle formulas are formulas used in finding the trigonometric values for angles that are doubled or halved.

    How are double angle and half angle formulas used?

    Double-angle and half-angle formulas are used by applying directly the formulas when finding double/half-angle trigonometric identities.

    What is an example of double-angle and half-angle formulas?

    An example of a double angle formula is sin2A = 2sinAcosA, while that of half-angle is sinA/2 = square root ((1-cosA)/2).

    How do you derive double angle formulas? 

    To derive double angle formulas you would need to apply the sum of trigonometric function formulas.

    What are the types of double angle formulas? 

    The types of double angle formulas are those of sin, cos, sec, cosec, tan and cot.

    Discover learning materials with the free StudySmarter app

    Sign up for free
    1
    About StudySmarter

    StudySmarter is a globally recognized educational technology company, offering a holistic learning platform designed for students of all ages and educational levels. Our platform provides learning support for a wide range of subjects, including STEM, Social Sciences, and Languages and also helps students to successfully master various tests and exams worldwide, such as GCSE, A Level, SAT, ACT, Abitur, and more. We offer an extensive library of learning materials, including interactive flashcards, comprehensive textbook solutions, and detailed explanations. The cutting-edge technology and tools we provide help students create their own learning materials. StudySmarter’s content is not only expert-verified but also regularly updated to ensure accuracy and relevance.

    Learn more
    StudySmarter Editorial Team

    Team Math Teachers

    • 12 minutes reading time
    • Checked by StudySmarter Editorial Team
    Save Explanation Save Explanation

    Study anywhere. Anytime.Across all devices.

    Sign-up for free

    Sign up to highlight and take notes. It’s 100% free.

    Join over 22 million students in learning with our StudySmarter App

    The first learning app that truly has everything you need to ace your exams in one place

    • Flashcards & Quizzes
    • AI Study Assistant
    • Study Planner
    • Mock-Exams
    • Smart Note-Taking
    Join over 22 million students in learning with our StudySmarter App
    Sign up with Email

    Get unlimited access with a free StudySmarter account.

    • Instant access to millions of learning materials.
    • Flashcards, notes, mock-exams, AI tools and more.
    • Everything you need to ace your exams.
    Second Popup Banner