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    Lesson 3 - Equivalent Expressions

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    iminathi ndlovu
    1) The document discusses equivalent expressions for trigonometric functions of the form y = a cos x + b sin x. 2) It shows that such expressions can be written as r sin(x + α) or r cos(x - α) by using trigonometric identities. 3) Examples are provided to demonstrate how to find the maximum and minimum values of such expressions and solve related equations.

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    Lesson 3 - Equivalent Expressions

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    iminathi ndlovu
    45 views15 pages
    1) The document discusses equivalent expressions for trigonometric functions of the form y = a cos x + b sin x. 2) It shows that such expressions can be written as r sin(x + α) or r cos(x - α) by using trigonometric identities. 3) Examples are provided to demonstrate how to find the maximum and minimum values of such expressions and solve related equations.

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    1) The document discusses equivalent expressions for trigonometric functions of the form y = a cos x + b sin x. 2) It shows that such expressions can be written as r sin(x + α) or r cos(x - α) by using trigonometric identities. 3) Examples are provided to demonstrate how to find the maximum and minimum values of such expressions and solve related equations.

    Copyright:

    © All Rights Reserved
    Download as PPT, PDF, TXT or read online from Scribd
    Download as ppt, pdf, or txt
    45 views15 pages

    Lesson 3 - Equivalent Expressions

    Uploaded by

    iminathi ndlovu
    1) The document discusses equivalent expressions for trigonometric functions of the form y = a cos x + b sin x. 2) It shows that such expressions can be written as r sin(x + α) or r cos(x - α) by using trigonometric identities. 3) Examples are provided to demonstrate how to find the maximum and minimum values of such expressions and solve related equations.

    Copyright:

    © All Rights Reserved
    Download as PPT, PDF, TXT or read online from Scribd
    Download as ppt, pdf, or txt
    of 15

    Core 4

    Trigonometry
    Equivalent Expressions
    Equivalent expressions for
    “y = a cos x + b sin x” Link
    E.g. y = 3 sin x + 4 cos x

    Period

    Amplitude

    y = a cos x y = b sin x

    The sine and cosine curves combine to make a new trig. curve that:
    • has a amplitude (max height) that is not 1
    • a period that is not 360 (2)
    • transformed left/right of normal sine and cosine curves
    Using Equivalent Expressions
    E.g. y = 3 sin x + 4 cos x
    Recall: sin (A + B) = sin A cos B + cos A sin B
    sin (x + ) = sin x cos  + cos x sin 
    sin (x + ) = cos  sin x + sin  cos x
    cos  and sin  cannot be bigger than 1, to get the 3 and the 4 in
    y = 3 sin x + 4 cos x
    We need a scaler (r):
    r sin (x + ) = (r cos ) sin x + (r sin ) cos x
    3 sin x + 4 cos x
    r cos  = 3 r sin  = 4
    r and  are values that can be found be simultaneous equations
    Example 1:
    Find the maximum value of:: y = 3 sin x + 4 cos x
    r sin (x + ) = (r cos ) sin x + (r sin ) cos x
    r cos  = 3 r sin  = 4
    r2 cos2  = 9 r2 sin2  = 16
    r2 cos2  + r2 sin2  = 9 + 16 = 25
    r2 (cos2  + sin2 ) = 25
    Since, cos2  + sin2  = 1
    r2 = 25
    r = 25 = 5
    Example 1: (cont)
    Find the maximum value of:: y = 3 sin x + 4 cos x
    r sin (x + ) = (r cos ) sin x + (r sin ) cos x
    r cos  = 3 r sin  = 4
    We found, r = 5
    To find  : or : r sin  = 4
    r cos  = 3 Either way : r cos  = 3
     = 53.1o r sin  = 4
    5 cos  = 3 r cos  3
    cos  = 3/5 tan  = 4/3
    Hence, y = 3 sin x + 4 cos x = 5 sin (x + 53.1)
    The maximum value is when sin (x + 53.1)=1. So the maximum is 5
    Example 2:
    Solve: 3 sin x + 4 cos x = 2
    In the range 0o < x < 360o
    Using the previous steps,
    3 sin x + 4 cos x = 5 sin (x + 53.1)
    The equation becomes ….
    5 sin (x + 53.1) = 2
    sin (x + 53.1) = 2/5
    x + 53.1o = 26.6o or x + 53.1o = 153.4o
    x = -26.5o or x = 100.3o
    x = 333.5o
    Task
    Page 63

    B8
    A) Expand r sin (x - )

    Using, sin (A - B) = sin A cos B - cos A sin B


    r sin (x - ) = r cos  sin x - r sin  cos x
    r sin (x - ) = (r cos ) sin x - (r sin ) cos x
    B) Find r for:: 4 sinx - 3 cos x = r sin (x - )
    r cos  = 4 r sin  = 3
    r2 cos2  + r2 sin2  = 16 + 9 = 25
    r2 (cos2  + sin2 ) = 25
    Since, cos2  + sin2  = 1
    r2 = 25
    r = 25 = 5
    r sin (x - ) = (r cos ) sin x - (r sin ) cos x = 4 sinx - 3 cos x
    C) Show that tan  = 0.75
    r cos  = 4 r sin  = 3
    r sin  = 3
    r cos  = 4
    r sin  = 3
    r cos  4
    tan  = 3/4 = 0.75
    D) Express 4 sinx - 3 cos x as r sin (x - )

    We found :: r=5 tan  = 0.75


     = tan-1(0.75)
     = 36.9o

    4 sinx - 3 cos x = 5 sin (x - 36.9)


    E) minimum value of ::: 4 sinx - 3 cos x

    minimum value of ::: 5 sin (x - 36.9)


    Minimum when sin (…) = -1
    Minimum = 5 x -1 = -5
    … at what value?

    sin (x - 36.9) = -1
    x - 36.9 = sin-1(-1) = 270o
    x = 270 + 36.9 = 306.9o
    E) minimum value of ::: 4 sinx - 3 cos x

    5
    4
    3
    2
    1
    0 Radians
    -1
    -2
    -3
    -4
    -5 Minimum is at -5

    About 5.3 radians


    [for min] x = 306.9o = 303o
    Using Equivalent Expressions
    General case for: y = a sin x + b cos x
    Using, sin (x + ) = cos  sin x + sin  cos x

    a sin x + b cos x can be expressed as: r sin (x + )


    Using, cos (x - ) = cos  cos x + sin  sin x
    a sin x + b cos x can be expressed as: r cos (x - )

    General case for: y = a sin x - b cos x


    Using, sin (x - ) = cos  sin x - sin  cos x

    a sin x - b cos x can be expressed as: r sin (x - )


    Using, cos (x + ) = cos  cos x - sin  sin x
    a sin x = b cos x can be expressed as: r cos (x + )
    Do : Now/ Friday
    Page 65: Exercise B
    Q3, Q4
    Page 67: Test Yourself
    Q12, 14
    Things you should know - 1
    y = sin x
    sin 0 = 0
    sin 90 =1

    sin x = cos (90-x)


    cos x = sin (90-x)
    y = cos x

    cos 0 = 1
    cos 90 =0
    Things you should know - 2

    sin 45 = 1/2 = 0.707


    1 2 cos 45 = 1/2 = 0.707
    45o
    1

    sin 30 =1/2= 0.5


    60o
    2
    sin 60 = /2 = 0.866
    3 3

    cos 60 =1/2= 0.5 30o


    1
    cos 30 = 3/2 = 0.866

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