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# Find the indicated trigonometric value in the specified quadrant. Function Quadrant IV, cot θ = -2, trigonometric Value sin θ. Thanks!

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Find the indicated trigonometric value in the specified quadrant. Function Quadrant IV, cot θ = -2, trigonometric Value  sin θ. Thanks!

Sep 13, 2017

#1
+7348
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cot θ  =  -2

By the definition of cotangent, this means....

$$\frac{\cos\theta}{\sin\theta}=-2$$

Multiply both sides of the equation by  sin θ .

$$\cos\theta=(-2)\sin\theta$$

The Pythagorean identity says....

(sin θ)2 + (cos θ)2  =  1

Plug in   (-2) sin θ   for   cos θ  .

(sin θ)2 + ( (-2) sin θ )2  =  1

Mulltiply out the exponents.

sin2θ + 4 sin2θ  =  1

Combine like terms.

5 sin2θ  =  1

Divide both sides by  5 .

sin2θ  =  1/5

Since sin is negative in Quad IV, Take the negative sqrt of both sides.

sin θ  =  $$-\sqrt{\frac15}\,=\,-\frac{\sqrt1}{\sqrt5}\,=\,-\frac{1}{\sqrt5}\,=\,-\frac{\sqrt5}{5}$$

*edit*

Also note that the only Quadrants that have an angle with a negative cotangent are II and IV, so it would be impossible for the angle to be in Quad I or III .

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Sep 13, 2017
edited by hectictar  Sep 13, 2017
edited by hectictar  Sep 13, 2017
#2
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Much easier and faster to realise that if cot (x)  = -2  then

tan (x) = -1/2 , construct the triangle and read off the value of sin (x)

----construct the triangle in the 4 th quadrant where the side  of the triangle on the x- axis  has length 2

hypoteuse has length Sqrt5 and opposite side to angle (x) has length  - 1             ie this means that tan (x) = -1/2

Now you can just read off the value by looking at the triangle ,  sin (x) =  - sqrt (1/5)  .  And it's also obvious from the diagram that sin (x) has to be negative.....

Sep 13, 2017
#3
+7348
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How did you know to make the hypotenuse sqrt5  ?

hectictar  Sep 13, 2017