Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 month ago

Given tan(𝜃)=− square root 7/2 , with sec(𝜃)>0, find sin(𝜃)=?

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  • ?
    Lv 7
    1 month ago

    tan(A)=-sqr(7/2) or tan(A)=-sqr(7)/2? not clear.

  • 1 month ago

    tan𝜃 = -√7/2

    sec𝜃 > 0...i.e. 1/cos𝜃 > 0

    Hence, cos𝜃 > 0

    If tan𝜃 < 0 and cos𝜃 > 0...𝜃 is in Q4

    Now, sec²𝜃 = tan²𝜃 + 1  

    so, sec²𝜃 = (-√7/2)² + 1 => 11/4

    Hence, sec𝜃 = ±√11/2

    i.e. sec𝜃 = √11/2...as sec𝜃 > 0

    Then, cos𝜃 = 2√11/11

    Now, tan𝜃 = sin𝜃/cos𝜃

    i.e. -√7/2 = sin𝜃/(2√11/11)

    Hence, sin𝜃 = -√77/11

    Note: sin²𝜃 + cos²𝜃 = 1

    checking gives: 

    (-√77/11)² + (2√11/11)² => 77/121 + 44/121 = 1

    :)>

  • 1 month ago

    sec(θ) > 0

    1/cos(θ) > 0

    cos(θ) > 0 → given that tan(θ) < 0 → then: sin(θ) < 0

    tan(θ) = - √(7/2)

    sin(θ)/cos(θ) = - √(7/2)

    cos(θ) = - sin(θ)/√(7/2)

    cos²(θ) = [- sin(θ)/√(7/2)]²

    cos²(θ) = sin²(θ)/(7/2)

    cos²(θ) = sin²(θ) * (2/7)

    cos²(θ) = (2/7).sin²(θ)

    cos²(θ) + sin²(θ) = 1

    (2/7).sin²(θ) + sin²(θ) = 1

    [(2/7) + (7/7)].sin²(θ) = 1

    (9/7).sin²(θ) = 1

    sin²(θ) = 7/9

    sin(θ) = ± (√7)/3 → recall: sin(θ) < 0

    sin(θ) = - (√7)/3

  • Vaman
    Lv 7
    1 month ago

    tan theta=sin theta/cos theta= 7/sqrt(7^2+2^2)= 7/53. cos theta= 2/53. You can guess the other values.

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  • Jim
    Lv 7
    1 month ago

    my answer is "Anonymous" as your post is Anonymous. There is no need for an Anonymous question in this category.

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