Asked by emma
Find the exact value of (sinθ) (cosθ) if cosθ > 0 and tanθ = −2/3
Answers
Answered by
oobleck
Draw a right triangle with legs 2 and 3, in QIV. Then
y = -2
x = -3
r = √13
sinθ = y/r
cosθ = x/r
y = -2
x = -3
r = √13
sinθ = y/r
cosθ = x/r
Answered by
emma
I'm confused with this topic, can you please explain it some more
Answered by
oobleck
Really? Read about angles in the standard position.
tanθ < 0 in QII and QIV. You say cosθ is positive, so that means x is positive
Did you draw the triangle?
sinθ = y/r = -2/√13
cosθ = x/r = -3/√13
Note that sin^2θ + cos^2θ = 1, as required.
sin^2θ + cos^2θ = y^2/r^2 + x^2/r^2 = (x^2 + y^2)/r^2 = 1
tanθ < 0 in QII and QIV. You say cosθ is positive, so that means x is positive
Did you draw the triangle?
sinθ = y/r = -2/√13
cosθ = x/r = -3/√13
Note that sin^2θ + cos^2θ = 1, as required.
sin^2θ + cos^2θ = y^2/r^2 + x^2/r^2 = (x^2 + y^2)/r^2 = 1
Answered by
emma
OHH thank you so much, i understand it now
so then the exact value would just be 1 though?
so then the exact value would just be 1 though?
Answered by
oobleck
no! The exact value of sinθ cosθ = (-2/√13)(-3/√13) = 6/13
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