Download Lecture 23

Math 140
Since tan 6  sin 6 / cos 6    12 /
Product-to-sum rules
RECALL
cosx  y  cos x cos y  sin x sin y
cosx  y  cos x cos y  sin x sin y
Adding these gives
cosx  y  cosx  y  2 cos x cos y
Subtracting gives
cosx  y  cosx  y  2 sin x sin y
Similarly
sinx  y  sin x cos y  cos x sin y
sinx  y  sin x cos y  cos x sin y
Adding these gives
sinx  y  sinx  y  2 sin x cos y
RECALL. sin(x) and cos(x) are always between -1 and 1.
`cos x  3
iff never. tno solution.
`2 cos 2 x  cos x  1
2 cos 2 x  cos x  1  0
2 cos x  1cos x  1  0
cos x  12 or cos x  1
x = p/3 + 2pn or x = -p/3 + 2pn or x = p + 2pn.
Dividing by 2 gives
PRODUCT-TO-SUM FORMULA.
sin x sin y  12 [cosx  y  cosx  y ]
cos x cos y  12 [cosx  y  cosx  y ]
sin x cos y  12 [sinx  y  sinx  y ]
Write as a sum or difference of trigonometric functions.
`sin 2 cos 6
 12 sin 2  6   sin 2  6 
 12 sin

3
 sin
2
3 .
You could continue and get:
3 /2.
`cos x cos 2x
 12 cosx  2x  cosx  2x
 12 cosx  cos3x  12 cosx  cos3x.
Solving trigonometric equations
RECALL. sin x  sin  x, cos x  cosx.
Note: x and   x are called supplementary angles.
Find all solutions for each equation.
`sin   12
y=1/2
p-p/6=5p/6
3
2
  13
p/6 is one solution. Since tan(x) has period p, adding a
multiple of p also gives a solution. The general solution
is x = p/6 + pn.
For sin, cos, add 2pn to the (usually two) simplest solutions.
For tan,
add pn to the one simplest solution.
`sin 2x  1
iff 2x = p/2 + 2pn,
Only one simple solution here.
iff x = p/4 + pn.
Lecture 23
p/6
`sin 2 x  cos x  1  0
1  cos 2 x  cos x  1  0
 cos 2 x  cos x  2  0
cos 2 x  cos x  2  0
cos x  1cos x  2
cos x  1 or cos x  2The second is impossible.
x = p + 2pn.
`2 tan 2 x  3 tan x sec x  2 sec 2 x  0
2 sin 2 x  3 sin x  2  0
2 sin x  1sin x  2  0
sin x   12 or sin x  2 The second is impossible.
x = -p/6 + 2pn or x = -5p/6 + 2pn.
RECALL. A function is 1-1 iff no horizontal line crosses its
graph more than once.
sin, cos and tan are not 1-1. But they are 1-1 on the
heavily marked intervals. These are the largest such
intervals containing first quadrant angles.
sin
-p/2
The two simplest solutions: q = p/6, and q = p-p/6 = 5p/6.
Since sin(q) has period 2p, adding a multiple of 2p to
either of these also produces a solution.
The general solution consists of the two sets
q = p/6 + 2pn or q = 5p/6 + 2pn for n an integer.
`cos x  12
Here we use x for the angle instead of q.
The two simplest solutions are x = p/3 and x = -p/3.
The general solution is
x = p/3 + 2pn or x = -p/3 + 2pn for n an integer.
CONVENTION. Assume n is an arbitrary, possibly negative,
integer. We’ll omit the phrase “for n an integer”.
`tan x  13
p/2
cos
p
0
tan
-p/2
p/2
For sin, the interval is [-p/2, p/2].
For cos, the interval is [0, p].
For tan, the interval is (-p/2, p/2).