Trigonometric question?

If  sin⁡x+cos⁡x = -√2 

What is  sin^3 x+ cos^5 x=?

3 Answers

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  • Let's pretend we don't know that x must be 5pi/4 radians

    sin(x) + cos(x) = -sqrt(2)

    (sin(x) + cos(x))^2 = (-sqrt(2))^2

    sin(x)^2 + 2sin(x)cos(x) + cos(x)^2 = 2

    1 + sin(2x) = 2

    sin(2x) = 1

    2x = arcsin(1)

    2x = pi/2 + 2pi * k

    x = pi/4 + pi * k

    k is an integer.

    Just using basic trig identities, namely sin(x)^2 + cos(x)^2 = 1 and 2sin(x)cos(x) = sin(2x), we have possible values for x

    From 0 to 2pi we have 2 possible values: pi/4 and pi/4 + pi, or 5pi/4

    sin(pi/4) + cos(pi/4) = sqrt(2)/2 + sqrt(2)/2 = sqrt(2)

    So that's not it

    sin(5pi/4) + cos(5pi/4) = -sqrt(2)/2 - sqrt(2)/2 = -sqrt(2)

    Bingo

    sin(x)^3 + cos(x)^5 =>

    sin(5pi/4)^3 + cos(5pi/4)^5 =>

    (-sqrt(2)/2)^3 + (-sqrt(2)/2)^5 =>

    (-sqrt(2)/2)^3 * (1 + (-sqrt(2)/2)^2) =>

    (-2 * sqrt(2) / 8) * (1 + 2/4) =>

    (-sqrt(2)/4) * (1 + 1/2) =>

    (-sqrt(2)/4) * (3/2) =>

    -3 * sqrt(2) / 8

  • 1 month ago

    x = 225 degrees  (subtract cos x from both sides, square both sides, use sin^2 = 1 - cos^2, solve quadradic equation).

    sin x = -sqrt(2)/2, cos x = -sqrt(2)/2

    (-sqrt2/2)^3 = - 2sqrt2/8 = -sqrt(2)/4

    (-sqrt2/2)^5 = - 4sqrt2/32 = -sqrt(2)/8

    answer = - 3 sqrt(2)/8

  • 1 month ago

    If sin⁡x+cos⁡x = -√2  then sin x = cos x = -√2 / 2

    so x could be 5π/4

    not that it matters...

    (-√2/2)^3 = -√8 / 8 = -2√2 / 8 = -√2 / 4

    (-√2/2)^5 = -√32 / 32 = -4√2 / 32 = -√2 / 8 so add those

    imo anyway

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