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Any help would be greatly appreciated. I am completely clueless on all of this stuff. We recently got a new teacher and she doesn't know what she's doing, and we have a huge test tomorrow on this. These are some of the study guide questions that I am completely clueless on.

1.) Determine the quadrant in which Θ lies: tanΘ<0, cosΘ>0
2.) Given sinΘ= -1/5 and tan Θ<0, find cosΘ
3.) Find the reference angle for Θ=228°
4.) Use the reference angle to find the exact value of cot(-150°)
5.) Use the reference angle to find the exact value of csc(225°)
6.) Use the reference angle to find the exact value of sin(- 8π/3)
7.) Use the reference angle to find the exact value of sec(7π/4)
8.) Given cosΘ= -7/8 and π/2 ≤ Θ ≤ π, find tanΘ
9.) Given sinΘ= -4/7 and 3π/2 ≤ Θ ≤ 2π, find secΘ

Respuesta :

mrinconj
When a triangle is formed by a ray starting at the origin and making some angle with the x-axis, we let y-component be the opposite to the angle or the rise of the triangle and let the x-component be the adjacent of the angle or the run. Hypotenuse (r) is always positive.
QUESTION 1:
So for cos Ф ≥ 0 (sorry for the symbol) you have cos Ф=adjacent/hypotenuse so CosФ=x/r oviously x have to be positive because r is positive all the time and you need  equal symbols for ≥ 0.
And for tanФ≤0 we know tanФ=opposite/adjacent i mean tanФ=y/x you kno now that x is positive from above then y must be negative because for getting ≤0 the symbols have to be differents, so we have x positive and y negative we are in Quadrant IV.
QUESTION 2:
From Pythagorean we know r^2=adj^2+opp^2
we have sinФ=-1/5, sinФ=opp/r... so opp=y=-1 and r=5
adj^2=r^2-opp^2....  adj=root(5^2-(-1)^2)=root(24)=2root(6)=x
Now if tanФ≤0 we have tanФ=opp/adj=y/x. From senФ=-1/5 we know y=-1 so x must be positive because ≤0
Solve cosФ=adj/r=(2root(6))/5=0.98
QUESTION 3:
reference angle for Θ=228°
the reference angle It is the angle formed with the x axis in the quadrant where the line formed by the angle is given, in this case the angle is in quadrant III so you have to substract 228-180=48
QUESTION 4:
Draw the angle from positive x-axis down so for the reference add -150+180=30, from trig table we know cos(30)=root(3)/2 but the angle -150 is in quadrant III so cosФ=x/r then x must be negative so cos(-150)=-root(3)/2.
I hope that I can continue tomorrowif you want. Sorry is too late for me now.