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

Guest Sep 13, 2017
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3+0 Answers

 #1
avatar+5256 
+2

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 .

hectictar  Sep 13, 2017
edited by hectictar  Sep 13, 2017
edited by hectictar  Sep 13, 2017
 #2
avatar+71 
+1

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.....

frasinscotland  Sep 13, 2017
 #3
avatar+5256 
0

How did you know to make the hypotenuse sqrt5  ?

hectictar  Sep 13, 2017

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