Let a be an angle, with 0≤α<2π0≤α<2π. Given cos(2α)=17/49 and 2α is in quadrant IV,

Comment (1) on “Let a be an angle, with 0≤α<2π0≤α<2π. Given cos(2α)=17/49 and 2α is in quadrant IV,”

1. tan(theta) = y/x = 2/1 = 2

   sin(theta) = y/r = 2/√5

   cos(theta) = x/r = 1/√5

   tan(theta) = y/x = -2/√5

   sin(theta) = y/r = -2/-1 = 2

   cos(theta) = x/r = -1/√5

   Step-by-step explanation:

   -2x+y =0

   y = 2x

   So the ratio between y to x is:

   y/x = 2

   y:x = 2:1

   For example, if x = 1, y = 2. Using this we'll get the hypotenus or the length of the terminal side as below

   r = √(1^2 + 2^2) = √5

   So the trigo for these are:

   tan(theta) = y/x = 2/1 = 2

   sin(theta) = y/r = 2/√5

   cos(theta) = x/r = 1/√5

   Another 3 trigo is when both x and y is negative (3rd quadrant)

   Ex: x=-1, y=-2

   r will be √5 as well

   Hence

   tan(theta) = y/x = -2/√5

   sin(theta) = y/r = -2/-1 = 2

   cos(theta) = x/r = -1/√5

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