

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 socalled halfangle and doubleangle 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)
hawley
I 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
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Update: June 2012

