Trigonometric Equation: 1 + sin(x) = cos(x)?

1 + sin(x) = cos(x)

How can I solve this...none of the squaring stuff...my teacher won't have it!

Update:

Unfortunately, I cannot give a solution by inspection. Also, how does multiplying by (1-sinx) help at all? You end up with an even more complex equation: cosx(cosx+sinx-1)=0.

9 Answers

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  • 1 decade ago
    Best answer

    1 + sin x = cos x

    I suggest to express sin x and cos x as functions of tan(x/2):

    sin x = [2 tan(x/2)] / [1 + tan²(x/2)]

    cos x = [1 - tan²(x/2)] / [1 + tan²(x/2)]

    thus

    1 + sin x = cos x →

    1 + [2 tan(x/2)] / [1 + tan²(x/2)] = [1 - tan²(x/2)] / [1+ tan²(x/2)] →

    let us multiply all terms by [1 + tan²(x/2)]:

    [1 + tan²(x/2)] + [2 tan(x/2)] = [1 - tan²(x/2)] →

    1 + tan²(x/2) + 2 tan(x/2) = 1 - tan²(x/2) →

    1 + tan²(x/2) + 2 tan(x/2) - 1 + tan²(x/2) = 0 →

    2 tan²(x/2) + 2 tan(x/2) = 0 →

    2 tan(x/2) [ tan(x/2) + 1] = 0 →

    tan(x/2)₁= 0 → (x₁/2) = 0 + n180°→ x₁= n360°

    tan(x/2)₂= - 1 → (x₂/2) = 135°+ n 180°→ x₂= 270°+ n360°

    being n a relative integer

    Bye!

  • 1 decade ago

    Let Sin(x) = s. Then, squaring both sides, we have:

    1 + 2s + s² = 1 - s²

    2s(1+s) = 0

    which has roots s = 0, -1. In other words, 1 + Sin(x) = Cos(x) holds true when Sin(x) = 0, and Sin(x) = -1, or when x = 0 + n π, and x = (3/2) π + n π, where n is any positive or negative integer.

    The other way to do it is to just plot the functions out, and see how that the above is true. Maybe that's how your teacher wants you to do it, because doing this by "squaring stuff" is just too much excellence in math.

  • Sanra
    Lv 4
    4 years ago

    For the best answers, search on this site https://shorturl.im/GxbP3

    Combine trig functions, their periods and horizontal shifts are the same: sin(x) + cos(x) = 1 √(2) ⋅ sin(x + ¼π) = 1 sin(x + ¼π) = ½√(2) Or you could square both sides: sin(x) + cos(x) = 1 [ sin(x) + cos(x) ]² = 1² sin²(x) + 2sin(x)cos(x) + cos²(x) = 1 1 + 2sin(x)cos(x) = 1 2sin(x)cos(x) = 0 sin(2x) = 0 ...... or ....... sin(x) = 0; cos(x) = 0

  • 3 years ago

    1 Sinx

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  • John
    Lv 7
    1 decade ago

    This is one of those that you just have to think about and look at for a while.

    Obviously, x = 0 works since

    1 + sin(0) = cos(0) which is the same as (1 + 0 = 1.)

    As it turns out, x = 0 is the only solution.

    -John

  • Como
    Lv 7
    1 decade ago

    cos x - sin x = 1

    Let

    cos x - sin x = k cos (x - α )

    cos x - sin x = (k cos α) cos x + (k sin α) sin x

    1 = k cos α

    - 1 = k sin α----------- (where k > 0)

    tan α = - 1 (α in 4th quadrant)

    α = 315°

    1² + (-1)² = k²

    k² = 2

    k = √ 2

    cos x - sin x = √ 2 cos (x - 315°)

    √ 2 cos (x - 315°) = 1

    cos (x - 315°) = 1 /√ 2

    x - 315° = 45° or 315°

    x = 360° or 630°

    x = 360° or 270°

  • 5 years ago

    1+sin(x)/n

    1+sin(x)/n, cross out both n's

    1+six=7

  • 1 decade ago

    multiply both sides by (1- sin x) and take it from t here johnny boy

  • Anonymous
    6 years ago

    sin3x*sinx

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