Help i don’t understand my math answer?
Suppose that cos theta= 2/5 and that theta is in the 4th quadrant. Find sin theta and tan theta exactly.
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answers:
llaffer say: Quadrant 4 is the section on the grid that has a positive "x" value and a negative "y" value.

Recall that the cosine is adj/hyp, so we can plot a point where x = 2 with an unknown "y" (though we know it's negative) with a hypotenuse of 5.

Using pythagarean theorem, we can find this unknown "y":

a² + b² = c²
2² + b² = 5²
4 + b² = 25
b² = 21
b = √21

So the coordinate of the point is (2, -√21)

Knowing that we can solve for the sine of this angle would be opp/hyp, or √(21) / 5. But since this is Q4, sin is negative, so sin(θ) = -√(21)/5

For tangent, we have opp/adj or √(21)/2, and is also negative due to being Q4, so:

tan(θ) = -√(21)/2
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ted s say: the reference triangle for that angle is { 2 , -√21 , 5 }
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Como say: y² = 5² - 2² = 21

y = √21

sin ∅ = - √21/5 , tan ∅ = - √21/2
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khalil say: first. Θ is in quad 4 → sinΘ < 0 and cosΘ > 0

sin² Θ = 1 - cos² Θ

sin² Θ = 1 - (2/5)²

sinΘ = ± (√21)/5

sinΘ = - (√21) / 5 ....accepted answer ◄◄◄

tanΘ = sinΘ / cosΘ

tanΘ = [(-√21)/5] / (2/5) = - (√21) / 2 ◄◄◄
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Pinkgreen say: Let A=theta, cosA=2/5
=>
in the iv th quadrant,
sinA=
-sqr[1-cos^2A]=
-sqr[1-(2/5)^2]=
-sqr(21)/5

tanA=
-sqr[sec^2(A)-1]=
-sqr[(5/2)^2-1]=
-sqr(21)/2
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Johnathan say: With theta in quadrant 4, you know that sin(theta) < 0. To find it, use the Pythagorean theorem (or sin^2(theta) + cos^2(theta) = 1).

2^2 + x^2 = 5^2

4 + x^2 = 25

x^2 = 21 -> x = +/- sqrt(21).

That means sin(theta) = -sqrt(21) / 5; remember that sin(theta) < 0 in quadrants 3 and 4. Then tan(theta) = sin(theta)/cos(theta) = (-sqrt(21) / 5) / (2/5) = -sqrt(21) / 2.
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