SW-MO ARML Practice – 4/18/2010
“Exotic Trig”
Even though we go beyond SOH-CAH-TOA – don’t forget it! Use the unit circle concept to
approximate the sine and cosine of GIT and HIP . What about the tangent? Secant? Cosecant?
Label the locations for the special values around the unit circle in the positive direction. Give the cosine
and sine values for those in the first quadrant. Then list the special values around the unit circle in the
negative direction.
Give the equation of the unit circle:
This leads to the basic Pythagorean Identity:
From this, we can easily derive two more:
Sum/Difference Formulas:
Cosine: cos(   )  cos  cos 
sin  sin 
Sine: sin(   )  sin  cos   cos  sin 
Tangent: tan(   ) 
“Crunchy Cashew Chicken is a Springfield Sin”
“Successful Springfield Cardinals Can Swing”
tan   tan 
1 tan  tan 
Double-Angle Formulas:
Cosine: cos(2 )  cos2 ( )  sin 2 ( )  2cos 2 ( )  1  1  2sin 2 ( )
Sine: sin(2 )  2sin  cos 
Tangent: tan(2 ) 
2 tan( )
1  tan 2 ( )
Half-Angle Formulas:
1  cos( )
 
Cosine: cos    
2
2
1  cos( )
 
Sine: sin    
2
2
Cofunction Formulas:


sin   cos    
2



cos   sin    
2



tan   cot    
2



cot   tan    
2



csc   sec    
2



sec   csc    
2

Law of Sines:
sin A sin B sin C


a
b
c
Area of a Triangle: A 
Law of Cosines: c 2  a 2  b 2  2ab  cos C
bh
1
or A  ab  sin C or A  s(s  a)(s  b)(s  c)
2
2
Some problems to do WITH NO CALCULATOR. Most are stolen from Art of Problem Solving.
NOTE: If you are seeing a big mess, you have missed an elegant option!!
1. Find the exact value for sin(105)
 9 
2. Find the exact value for cos 

 8 
3. Evaluate: tan(10)  tan(20)  tan(30)  tan(40)  tan(50)  tan(60)  tan(70)  tan(80)
4. Write in the form asin(x°):
sin13  sin167  cos13  cos167sin13  sin167  cos13  cos167
5. Find x if arctan x  tan 1 (4)  tan 1 (6)
6. Given positive integer n and a number c, -1 < c < 1, for how many values of q in [0, 2π) is
sin(nq) = c ?
7. Given ABC , with side a opposite
and c = 5.
A (and so on), find sinA + sin2B + sin3C if a = 3, b = 4,
8. Compute the SMALLEST positive angle x such that: 8sin x cos x  8sin x cos x  1
5
5
9. If A = 20° and B = 25°, then find the value of (1 + tanA)(1 + tanB).
10. Find the area of the incircle of a triangle with side lengths 13, 14, and 15.
11. Find the length of the altitude to one of the legs in the triangle with sides: 10, 10, 12.
12. If the sides of a triangle are in the ratio 4:6:8, then find the cosine of the smallest angle.
13. In ABC , CD is the bisector of
1 1
 .
a b
 
C with D on AB . If cos C 2  13 and CD  6 , compute
“Exotic Trig” – Solutions
1. sin(105)  sin(60  45) 
 9
2. cos 
 8
6 2
4
 
1  cos 

 
4   2 2
   cos    
2
2

8
3. Turn the tangents into sines and cosines and recall the cofunction formula: sin   cos  90    ,
so everything cancels and you get 1.
4. Use the cofunction formulas and the fact that sin(180   )  sin  and cos(180   )   cos to
simplify the statement to:
 2cos 13  2sin 13  4sin 13 cos 13  2sin  26
5. Put each side of the equation into the tangent function as input. Use the sum formula for the
right side! 10
23
6. Since n is a positive integer, you will go around the unit circle an integer number of times, each
time finding two values, since c  1 , so there will be 2n solutions.
7. Sketch ABC . Note that 3-4-5 is a Pythagorean Triple, so this is a right triangle.
 3
double angle for sin(2B). For sin(3C), note that C is a right angle, so find sin 
 2
then becomes 14 .
25
Use the

 and the sum



5
5
4
4
8. Start by factoring the GCF: 8sin x cos x  8sin x cos x  1  8sin x cos x cos x  sin x  1
 8sin x cos x(cos2 x  sin 2 x)(cos2 x  sin 2 x)  1  4sin(2 x)cos(2 x) 1  1  2sin(4 x)  1


So, sin(4 x)  1 2 , 4 x  , x 
6
24
9. Note that 20 + 25 = 45. Let B = 45° – A. Then, since tan(45  A) 
 1  tan A 
(1  tan A)(1  tan B)  (1  tan A) 1 

 1  tan A 
1  tan A
 (1  tan A) 1  (1  tan A) 
 1  tan A  a  tan A  2
1  tan A
1  tan A
, we get:
1  tan A
10. Find the area of the triangle using Heron’s Formula. Then recall the formula for finding the area
of a regular polygon is ½ aP where a is the apothem (radius of the incircle) and P is the
perimeter of the polygon. This formula will also apply to triangles. A  16
11. Drop an altitude to the base of the isosceles triangle and find the area. Then consider a leg as
the new “base” and use the known area to get the length of the altitude. h = 9.6
12. Start by reducing the ratio to 2:3:4. Let the three sides be 2k, 3k, 4k and use the Law of Cosines.
Since the smallest angle is requested, set it up as:
 2k 
2
  3k    4k   2  3k  4k  cos  . The
2
2
k 2 ' s will cancel and you get cos  7 .
8
13. Set up the triangle as below. Add the areas of the two smaller triangle to equal the area of the
larger triangle.
 
A(ACD )  3b sin  
C
 2 
 
2 2
b
A
6
D
 
A(BCD )  3a sin  
 2
a
 1
cos( )=
2 3
B
A(ACB)  1 ab sin  
2
 
 
So, 3b sin    3a sin    1 ab sin  
2
 2
 2
 
 
 
 
And, 3b sin    3a sin    ab sin   cos  
 2 
 2 
 2 
 2 
Or,
ab
3b  3a ab
1 1 1
 
3b  3a  ab cos    3b  3a 


  
3
3ab
9ab
a b 9
 2 