Download Use the trig identity : sin C – sin D = 2 cos (C+D)

You need to refresh all your trig identities. There are plenty of web sites which talk
about those. I have high lighted the trig identities that I used in each problem.
1. Find the exact value of the sine, cosine, and tangent of the angle. 11pie/12
Use : sin x = sin (-x)
Sin (11/12) = sin(-11/12) = sin(/12)
Since we know what is sin pi/6 using the following identity makes sense.
Use: cos 2x = 1 – 2sin2 x 
sinx = sqrt {½ (1-cos2x)}
Sin (11/12) = sin(-11/12) = sin(/12) = sqrt {½ (1-cos(/6)}=  (1/2 (1-3/2))
Sin (11/12 = (2-3)/2 This is the exact value.
Evaluating we get Sin (11/12) = 0.258819
2. Find all solutions of the equation in the interval [0, 2pie).
Sin(x-5pie/6)-sin(x+ 5pie/6) =1
Use the trig identity : sin C – sin D = 2 cos (C+D)/2 * sin (C-D)/2
Sin(x-5pie/6)-sin(x+ 5pie/6) = 2 cos x * sin –(5pi/6) = 1
2 cos x * sin –(5pi/6) = 1  cos x * sin 5pi/6 = - 1/2,
we know sin 5pi/6 = sin pi – 5pi/6 = sin pi/6 = ½
cos x * 1/2 = -1/2, 
cos x = -1
Between 0 and 2pi we have only one solution which is x = pi.
3. Find the exact value of sin2x using the double angle formula. Please help explain
the use of the double angle formula.
Sin x=1/7, 0<x<pie/2
Double angle formula : sin 2x = 2 sin x * cos x
Sinx = 1/ 7 and cos x =  (1-sin2 x) =  ( 1- 1/49) =  (48) / 7 = 43/7
So, sin 2x = 2 sin x * cos x = 2 * 1/7 * 43/7 = 83/49
Sin 2x = 83/49 (exactly)
Evaluating sin 2x = 0.282784
4. Given cos theta = 4/9, where 3pie/2 <or equal to theta <or equal to 2pie.
Question incomplete??
5. Express 2sin3xcos6x as a sum containing only sines or cosines.
There are 4 identities involving sin C  sin D and cos C  cos D. These are very
useful. We will use one of them. (note we used one in prob 2)
Identity used: sin C + sin D = 2 sin (C+D)/2 * cos (C-D)/2
2sin3xcos6x = 2 sin (C+D)/2 * cos (C-D)/2 ( we will find C and D)
(C+D)/2 = 3x  C+D = 6x, (C-D)/2 = 6x  C-D = 12 x
Solving for C and D we get C = 9x, D = -3x
So we can write 2sin3xcos6x = sin 9x + sin (-3x ) = sin 9x + (-sin 3x)
2sin3xcos6x = sin 9x + sin (-3x ) = sin 9x + (-sin 3x)
6. Express cos7x-cos5x as a product containing only sines and/or cosines.
Similar to 5.
Identity used: cos C – cos D = 2 sin (C+D)/2 * sin (D-C)/2
cos7x-cos5x = 2 sin(6x)sin(-x) = -2 sinx sin 6x
7. Evaluate the expressions without the aid of a calculator.
a. Arctan(- sqrt3/3)
b. Arcsin(-1/2)
Assume x = tan –1 (-3/3) = tan –1 -(1/3)  tan x = -1/3
tan 2 x = 1/3  use the fact that 1 + tan 2 x = sec 2 x
sec 2 x = 1 + 1/3 = 4/3
sec x = 1/cos x = 2 / 3  cos x = 3 / 2 (familiar ?)
positive solution : cos x = 3 / 2  x= pi/6
cosine is positive in the first and forth quadrant. But we are interested only a
solution in the 2nd and 4th quadrant because tan is negative in these quadrants. (If
you see the original expression the given inverse tan function has a negative value
and tan function are negative only in the 2nd and 4th quads )
So one solution is
x = 2pi -pi/6 = 11 pi/6
negative solution : cos x = -3 / 2 
cosine is negative in the 2nd and 3rd quadrant. So solution could be x = pi – pi/6
and/or pi + pi/6, ie. X = 5pi/6, 7pi/6. But we are interested only a solution in the
2nd and 4th quadrant
Other solution is 5pi/6
Arctan(- sqrt3/3) = 5pi/6 or 11pi/6
(b) Arcsin(-1/2) = x
Sin x = -1/2  x = -pi/6
Sin is negative in 3rd and 4th quads. So x = pi+pi/6 = 7pi/6 and
X = 2pi – pi/6 = 11pi/6
Arcsin(-1/2) = 7pi/6 or 11pi/6
8. Use inverse functions to evaluate the expressions,
a. cos(arcsin(sqrt5/5)) ,
say x= arcsin(sqrt5/5)) 
sin x = sqrt5/5  sin x = 1/ 5
cos(arcsin(sqrt5/5)) = cos x = ?
Use a right triangle to find cosx which is what we need. sin x = 1/ 5 =
opposite side / hypotenuse.
Cos x = adjacent /hypo. = 2/5
b. cos(arctan(sqrt2/x))
As before, say, r = arctan(sqrt2/x))  tan r = 2/x,
hypot of the right triangle = (x2 + 2)
So cos(arctan(sqrt2/x)) = cos r adjacent/ hypot. = x/(x2 + 2)
9.
= Identify he x-values that are solutions of the equations.
a. 8cos x-4 = 0
b. 18cot^2 x-18 = 0
8cos x-4 = 0 
angles.
cos x = ½,
x = pi/3, 2pi- pi/3 = 5pi/3, cosine is cyclic in 2pi
General solutions are x = pi/3  2npi
and x = 5pi/3  2npi
where n = integer
18cot^2 x-18 = 0  cot^2 x = 1  tan^2 x = 1
sec2 x = 1 +tan^2 x = 2
cos^2 x = ½
cos x =  ½
positive: cos x = ½  similar to part (a)
Solution set 1: x = pi/3  2npi and x = 5pi/3  2npi
where n = integer
negative cos x = -1/2  x = pi  pi/3
Solution set 2: 4pi/3  2n pi or
2pi/3 2n pi , n= integer
10. A large pole is 175 feet tall. On a particular day at noon it casts a 198-foot
shadow. What is the sun’s angle of elevation?
Tan x = 175/198
Elevation = 41.5o