Download Trigonometry Quesitons (with Solutions)

SW-MO ARML Practice – 11/17/2013
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?
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 ( )  2cos2 ( ) 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
sin 
   1  cos 
Tangent: tan   

sin 
1  cos 
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  b2  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
A (and so on), find sinA + sin2B + sin3C if a = 3, b = 4,
and c = 5.
8. Compute the SMALLEST positive angle x (in radians) such that: 8sin x cos5 x  8sin 5 x cos x  1
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
 
C with D on AB . If cos C 2  13 and CD  6 , compute
1 1
 .
a b
14. The sides of a regular pentagon are extended to form a five-pointed star. If the ratio of the area
of the pentagon to the area of the star equals sin  , for 0    90 , compute the value of  .
15. Compute the smallest positive angle x, in degrees, such that
tan 4 x 
cos x  sin x
cos x  sin x
16. The diagonals of rhombus ABCD are 10 and 4. Rhombus CDEF is drawn in the same plane,
with DE  DA . Compute the area of rhombus CDEF.
11-17-2013 – 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  sin2 x)(cos2 x  sin2 x)  1  4sin(2x)cos(2x) 1  1  2sin(4x)  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
14. Without loss of generality, let DC = 2, making
DH = 1. Since the area of ΔDHG is 1/10 the
area of the pentagon and the area of ΔDFG is
1/10 the area of the star, and since DC  GF ,
A( pentagon) A  DHG  GH
GH



A(star )
A  DFG  GF GH  HF
Using SOH-CAH-TOA, we get:
sin 54
GH
tan 54
cos 54


GH  HF tan 54  tan 72 sin 54  sin 72
cos 54 cos 72
sin 54 cos 72
sin 54 cos 72


sin 54 cos 72  sin 72 cos 54
sin126
sin 54 cos 72

 cos 72  sin18
sin 54
Answer: 18°
sin 4 x cos x  sin x

. Cross multiply and look for sum formulas. You get to
cos 4 x cos x  sin x
tan 5x  1, so x = 9°
15. Hint: Start with:
16. Let m BCD   . The side of each rhombus is
A(CDEF) = 29  29  sin  90     29cos 
Using the Law of Cosines,
cos  
29  29  16 21

2  29
29
Then, A(CDEF) = 29 
21
 21
29
29 . Then,