Contest math trigonometry problems ```Name: julia c lipman Status: N/A Age: N/A Location: N/A Country: N/A Date: Around 1995 ``` Question: I am going to try out for the ARML math team in a few weeks and I have noticed a type of problem which seems to appear on the tryouts with some regularity. The problem will require the evaluation of an expression with trigonometry functions such as (sin 25)(sin 35)(sin 85). We are not allowed to use calculators and have a time limit. Here is another example: (tan 54)(cos 54 + cos 162 + cos 270) All angles are in degrees. Can you suggest any method for solving problems like these? Replies: Remember the basic trigonometry identities: ```sin^2(x) + cos^2(x) = 1 sin( a +/- b ) = sin(a)*cos(b) +/- sin(b)*cos(a) cos( a +/- b ) = cos(a)*cos(b) -/+ sin(a)*sin(b) ``` From these you can derive the so-called half-angle and double-angle formu- las: ```sin(a/2) = SQRT( (1 - cos(a))/2 ) cos(a/2) = SQRT( (1 + cos(a))/2 ) sin(2a) = 2*sin(a)*cos(a) cos(2a) = 1 - 2*sin^2(a)``` Of course, you need to remember the sine and cosine of the "easy" angles: 30, 45, 60, and 90 degrees. Note in your examples: 25 + 35 = 60, 54 = 3 * 18, 18 = 1/5 * 90, etc. If you know a little about complex numbers, you can derive ALL the fundamen- tal trigonometry identities from De Moivre's Theorem: e^(i * x) = cos(x) + i*sin(x) where i = SQRT(-1) hawleyI needed a few more identities, derivable from or special cases of those given by hawley, in the derivations below: ```sin(a)sin(b) = [cos(a - b) - cos(a + b)]/2 ; cos(a)sin(b) = [sin(a + b) - sin(a - b)]/2 ; cos(a) + cos(b) = 2*cos[(a + b)/2]*cos[(a - b)/2]; sin(90 + x) = sin(90 - x); sin(180 + x) = -sin(x); cos(180 - x) = -cos(x); sin(x) = cos(90 - x); cos(-x) = cos(x); sin(-x) = -sin(x) sin(25)sin(35)sin(85) = {[cos(-10) - cos(60)]/2}sin(85) = [cos(10)sin(85) - (1/2)sin(85)]/2 = {[sin(95) - sin(-75)]/2 - [sin(85)]}/2 = [sin(95) + sin(75) - sin(85)]/4 = sin(75)/4 = cos(15)/4 ``` Now you need the half angle identity for cosine, and the knowledge that cos(30) = sqrt(3)/2. The final result is sqrt([1 + {sqrt(3)/2}]/2)/4, or sqrt[2+sqrt(3)] / 8. ```tan(54)*[cos(54) + cos(162)+cos(270)] = sin(54) + [sin(54)cos(162)]/cos(54) + 0 = sin(54) + (1/2)[sin(216) - sin(108)]/cos(54) = sin(54) + [-sin(36)]/(2cos(54) - [2sin(54)cos(54)]/[2cos(54)] = -sin(36)/[2cos(54)] = -cos(54)/[2cos(54)] = -1/2 ``` Good knowledge of the identities is crucial, but as you can see there is a need for a certain amount of resourcefulness that can only come with practice. (I can tell you that it took me an embarrassingly long time to figure these out!) Good luck to you! rcwinther Click here to return to the Mathematics Archives

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