document.write( "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) \n" ); document.write( "
Algebra.Com's Answer #519695 by Theo(13342)![]() ![]() You can put this solution on YOUR website! you can solve this with a calculator or without. \n" ); document.write( "there are 2 basic methods that i know of. \n" ); document.write( "the first is to find the angles involved and then subtract them from each other ad then find the sin. \n" ); document.write( "the other is to use the trigonometric identities to find the answer. \n" ); document.write( "i'll do both. \n" ); document.write( "the answers should be the same. \n" ); document.write( "if sec(x) = sqrt(5)/2, then cos(x) = 2/sqrt(5). \n" ); document.write( "if cos(x) = 2/sqrt(5), then x = arccos(2/sqrt(5)) = 26.56... degrees. \n" ); document.write( "this is if the angle was in quadrant 1. \n" ); document.write( "since the angle is in quadrant 4, the angle would be 360 - 26.56... degrees which is equal to 333.43... degrees. \n" ); document.write( "the actual number of degrees is stored in my calculator. \n" ); document.write( "that gets us angle x. \n" ); document.write( "if tan(y) = -1/3, then y = arctan(-1/3) = -18.43... degrees. \n" ); document.write( "to make this angle positive, add 360 to it to get 341.56... degrees. \n" ); document.write( "this angle is in Q4. \n" ); document.write( "the equivalent reference angle in Q1 would be equal to 360 - 341.56... degrees which is equal to 18.43... degrees. \n" ); document.write( "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. \n" ); document.write( "this angle is also stored in the calculator. \n" ); document.write( "we get: \n" ); document.write( "angle x = 333.43... degrees \n" ); document.write( "angle y = 161.56... degrees. \n" ); document.write( "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. \n" ); document.write( "you can then transpose to Q2 by taking 180 - the angle in Q1. \n" ); document.write( "either way, you wind up with angle x = 333.43... and angle y = 161.56... \n" ); document.write( "the full version of each of these angles is stored in calculator memory. \n" ); document.write( "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. \n" ); document.write( "i get sin(x-y) = sin(333.43... - 161.56...) = sin(171.86...) = .1414213562.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "to do this without using the calculator requires the use of the trigonometric identity formulas.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the formula to use is sin(x-y) = sin(x)cos(y)-cos(x)sin(y). \n" ); document.write( "in order to use this formula, you have to find sin(x) and cos(y) and cos(x) and sin(y). \n" ); document.write( "you are given that sec(x) = sqrt(5)/2 with x in Q4 and tan(y) = -1/3 with y in Q2. \n" ); document.write( "you can use these relationships to find sin(x) and cos(x) and sin(y) and cos(y).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "start with sec(x) = sqrt(5)/2 with x in Q4. \n" ); document.write( "since sec(x) = 1/cos(x), you get 1/cos(x) = sqrt(5)/2 which results in cos(x) = 2/sqrt(5). \n" ); document.write( "this means the adjacent side of the angle is 2 and the hypotenuse is sqrt(5). \n" ); document.write( "use the pythagorean formuls of a^2 + b^2 = c^2 to find the opposite side. \n" ); document.write( "in this formula, assign a = 2 and c = sqrt(5) and solve for b. \n" ); document.write( "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. \n" ); document.write( "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). \n" ); document.write( "so far you have: \n" ); document.write( "cos(x) = 2/sqrt(5) \n" ); document.write( "sin(x) = -1/sqrt(5)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "start with tan(y) = -1/3 with y in Q2. \n" ); document.write( "when y is in Q2, the adjacent side is negative and the opposite side is positive. \n" ); document.write( "this means that tan(y) = 1/-3 because opposite side is positive and adjacent side is negative. \n" ); document.write( "use the pythagorean formula again to get a^2 + b^2 = c^2 \n" ); document.write( "assign 1 to a and -3 to b and you get: \n" ); document.write( "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. \n" ); document.write( "now you can find sin(y) and cos(y). \n" ); document.write( "sin(y) = 1/sqrt(10) \n" ); document.write( "cos(y) = -3/sqrt(10).\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "you now have all the ingredients you need. \n" ); document.write( "you have: \n" ); document.write( "sin(x) = -1/sqrt(5) \n" ); document.write( "cos(x) = 2/sqrt(5) \n" ); document.write( "sin(y) = 1/sqrt(10) \n" ); document.write( "cos(y) = -3/sqrt(10)\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the formula is sin(x-y) = sin(x)cos(y) - cos(x)sin(y) \n" ); document.write( "this formula becomes: \n" ); document.write( "sin(x-y) = (-1/sqrt(5))(-3/sqrt(10)) - (2/sqrt(5))(1/sqrt(10)) \n" ); document.write( "this simplifies to: \n" ); document.write( "sin(x-y) = 3/sqrt(50) - 2/sqrt(50) which results in: \n" ); document.write( "sin(x-y) = 1/sqrt(50) which has a decimal equivalent of .1414213562.\r \n" ); document.write( " \n" ); document.write( "\n" ); document.write( "the results are the same whether you use method 1 or method 2. \n" ); document.write( "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. \n" ); document.write( "a picture of what your angles look like is shown below: \n" ); document.write( " ![]() \n" ); document.write( "\n" ); document.write( " \n" ); document.write( " |