Math 1720 - Trigonometry
University of Memphis
Instructor: Gábor Mészáros
Central European University, Budapest
10.30.2013
Solutions
Recall the identities:
α+β
α−β
) cos(
)
2
2
α−β
α+β
sin α − sin β = 2 sin(
) cos(
)
2
2
1. Solve the following trigonometric equations on [0, 2π]!
sin α + sin β = 2 sin(
(1)
(2)
(a) sin x + sin(x + π4 ) = 0,
Using identity 1) we can transform our equation to
2 sin(
x+x+
2
π
4
) cos(
x − (x + π4 )
)=0
2
(3)
A product is 0 if and only if at least one component is 0...
x+3x
(b) sin x = sin(3x), Use the second identity, that is, sin x − sin(3x) = 2 sin( x−3x
2 ) cos( 2 ) = 0.
A product is 0 if and only if at least one component is 0...
(c) sin(2x) = sin(3x)
Use the same technique as above.
(d) sin x = sec x.
cos x 6= 0. sin x =
1
cos x ,
that is, sin x cos x = 1, 2 sin x cos x = 2, sin(2x) = 2, no solution.
2. Solve the following trigonometric equations on [0, 2π]! Use calculator (but not equation-solver!)
if necessary.
√
√
12 + ( 3)2 :
√
1
3
2
q
sin x + q
cos x = q
√
√
√ .
12 + ( 3)2
12 + ( 3)2
12 + ( 3)2
(a) sin x + 3 cos x = 2, Divide both sides by
That is,
Switch
1
2
to cos π3 and
√
3
2
√
1
3
sin x +
cos x = 1.
2
2
(4)
(5)
to sin π3 . It gives
cos
Which we can solve.
q
π
π
sin x + sin cos x = 1
3
3
π
sin(x + ) = 1
3
1
(6)
(7)
(b) 3 sin x + 4 cos x = −5,
√
Use the same technique as above and divide by 32 + 42 = 5. Use calculator to nd β for
such that cos x = 35 (out of the two possible choices choose the one for which sin x = 45
instead of 54 ), that is, x ≈ 0.64. Solve sin(x + 0.64) = 1.
(c)
√
3 sin x − cos x =
√
2. Use the same technique as above.
3. Solve the equation sin x + sin(2x − π2 ) = 2.
π
π
x−2x− 2
2
Transform to sin( x+2x−
) cos(
) = 2, that is,
2
2
sin(
x + 2x −
2
π
2
) cos(
π
2
x − 2x −
2
)=1
(8)
Observe that the product is upperly bounded by 1 and is equal to it if and only if
sin(
or
sin(
x + 2x −
2
x + 2x −
2
π
2
π
2
) = cos(
) = cos(
x − 2x −
2
x − 2x −
2
π
2
π
2
)=1
(9)
) = −1
(10)
In both cases solve the equations and look for common solutions.
2