Question 856113: if sec x = square root 5/2 with angle x in quadrant IV and tan y= -1/3 with angle y in quadrant II, find the value of sin (x-y)
Found 2 solutions by stanbon, Theo:
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! if sec x = square root 5/2 with angle x in quadrant IV and tan y= -1/3 with angle y in quadrant II, find the value of sin (x-y)
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Note: x in QIV where x is positive and y is negative
If sec(x) = sqrt(5)/2
cos(x) = 2/sqrt(5)
sin(x) = sqrt(5-4)/5 = -1/5
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Note y in QII where x is negative and y is positive
If tan(y) = -1/3
x = -3 ; y = 1 ; so r = sqrt[3^2+1] = sqrt(10)
sin(y) = 1/sqrt(10)
cos(x) = -3/sqrt(10)
tan(x) = -1/3
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sin(x-y) = sin(x)cos(y)-cos(x)sin(y)
sin(x-y) = (-1/5)(-3/sqrt(10))-(2/sqrt(5))(1/sqrt(10))
= (3-2sqrt(5))/sqrt(10)
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Cheers,
Stan H.
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Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! you can solve this with a calculator or without.
there are 2 basic methods that i know of.
the first is to find the angles involved and then subtract them from each other ad then find the sin.
the other is to use the trigonometric identities to find the answer.
i'll do both.
the answers should be the same.
if sec(x) = sqrt(5)/2, then cos(x) = 2/sqrt(5).
if cos(x) = 2/sqrt(5), then x = arccos(2/sqrt(5)) = 26.56... degrees.
this is if the angle was in quadrant 1.
since the angle is in quadrant 4, the angle would be 360 - 26.56... degrees which is equal to 333.43... degrees.
the actual number of degrees is stored in my calculator.
that gets us angle x.
if tan(y) = -1/3, then y = arctan(-1/3) = -18.43... degrees.
to make this angle positive, add 360 to it to get 341.56... degrees.
this angle is in Q4.
the equivalent reference angle in Q1 would be equal to 360 - 341.56... degrees which is equal to 18.43... degrees.
since the angle is in Q2, the equivalent angle in Q2 is equal to 180 - 18.43... degrees which is equal to 161.56... degrees.
this angle is also stored in the calculator.
we get:
angle x = 333.43... degrees
angle y = 161.56... degrees.
the easier way to find the equivalent angle in Q1 is to simply make tan(-1/3) equal to tan(1/3). this will automatically get you the angle in Q1.
you can then transpose to Q2 by taking 180 - the angle in Q1.
either way, you wind up with angle x = 333.43... and angle y = 161.56...
the full version of each of these angles is stored in calculator memory.
now that i know the angles, i should be able to find sin(x-y) by subtracting y from x and then finding the sine.
i get sin(x-y) = sin(333.43... - 161.56...) = sin(171.86...) = .1414213562.
to do this without using the calculator requires the use of the trigonometric identity formulas.
the formula to use is sin(x-y) = sin(x)cos(y)-cos(x)sin(y).
in order to use this formula, you have to find sin(x) and cos(y) and cos(x) and sin(y).
you are given that sec(x) = sqrt(5)/2 with x in Q4 and tan(y) = -1/3 with y in Q2.
you can use these relationships to find sin(x) and cos(x) and sin(y) and cos(y).
start with sec(x) = sqrt(5)/2 with x in Q4.
since sec(x) = 1/cos(x), you get 1/cos(x) = sqrt(5)/2 which results in cos(x) = 2/sqrt(5).
this means the adjacent side of the angle is 2 and the hypotenuse is sqrt(5).
use the pythagorean formuls of a^2 + b^2 = c^2 to find the opposite side.
in this formula, assign a = 2 and c = sqrt(5) and solve for b.
you will get 2^2 + b^2 = (sqrt(5)^2 which results in 4 + b^2 = 5 which results in b^2 = 1 which results in b = 1.
since a is the adjacent side and c is the hypotenuse, b must be the opposite side and sin(x) = opposite / hypotenuse = 1/sqrt(5). since x is in Q4, the opposite side has to be negative, so sin(x) = -1/sqrt(5).
so far you have:
cos(x) = 2/sqrt(5)
sin(x) = -1/sqrt(5)
start with tan(y) = -1/3 with y in Q2.
when y is in Q2, the adjacent side is negative and the opposite side is positive.
this means that tan(y) = 1/-3 because opposite side is positive and adjacent side is negative.
use the pythagorean formula again to get a^2 + b^2 = c^2
assign 1 to a and -3 to b and you get:
1^2 + (-3)^2 = c^2 which results in 1 + 9 = c^2 which results in c^2 = 10 which results in c = sqrt(10) because the hypotenuse is always positive.
now you can find sin(y) and cos(y).
sin(y) = 1/sqrt(10)
cos(y) = -3/sqrt(10).
you now have all the ingredients you need.
you have:
sin(x) = -1/sqrt(5)
cos(x) = 2/sqrt(5)
sin(y) = 1/sqrt(10)
cos(y) = -3/sqrt(10)
the formula is sin(x-y) = sin(x)cos(y) - cos(x)sin(y)
this formula becomes:
sin(x-y) = (-1/sqrt(5))(-3/sqrt(10)) - (2/sqrt(5))(1/sqrt(10))
this simplifies to:
sin(x-y) = 3/sqrt(50) - 2/sqrt(50) which results in:
sin(x-y) = 1/sqrt(50) which has a decimal equivalent of .1414213562.
the results are the same whether you use method 1 or method 2.
the answer is that sin(x-y) = .1414213562 in decimal format, and sin(x-y) = 1/sqrt(50) in fraction format or what is sometimes referred to as exact format since the decimal format can sometimes be an approximate value only.
a picture of what your angles look like is shown below:
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