Solutions to Assignment 11 - Math

Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz
1. Trigonometric Identities
Prove the following trigonometric identities:
(a)
tan(x) − cot(x)
= sin(x) cos(x)
tan2 (x) − cot2 (x)
sin(x)
cos(x)
−
tan(x) − cot(x)
cos(x)
sin(x)
=
2
2
2
tan (x) − cot (x)
sin (x)
cos2 (x)
−
2
cos (x)
sin2 (x)
sin2 (x) − cos2 (x)
sin(x) cos(x)
=
sin4 (x) − cos4 (x)
sin2 (x) cos2 (x)
2
sin (x) − cos2 (x)
sin2 (x) cos2 (x)
=
sin(x) cos(x)
sin4 (x) − cos4 (x)
(sin2 (x) − cos2 (x)) sin(x) cos(x)
(sin2 (x) − cos2 (x))(sin2 (x) + cos2 (x))
sin(x) cos(x)
=
sin2 (x) + cos2 (x)
= sin(x) cos(x)
=
(b) tan(x) + cot(x) = sec(x) csc(x)
tan(x) + cot(x) =
sin(x)
cos(x)
+
cos(x)
sin(x)
sin2 (x) + cos2 (x)
sin(x) cos(x)
1
=
sin(x) cos(x)
1
1
=
cos(x)
sin(x)
=
= sec(x) csc(x)
1
Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz
(c)
sin(3x) + sin(7x)
= cot(2x)
cos(3x) − cos(7x)
The following two sum to product identities will be used
u+v
u−v
sin(u) + sin(v) = 2 sin
cos
2
2
u+v
u−v
cos(u) − cos(v) = −2 sin
sin
2
2
Thus we get the following:
3x + 7x
3x − 7x
cos
sin(3x) + sin(7x)
2
2
=
3x + 7x
3x − 7x
cos(3x) − cos(7x)
−2 sin
sin
2
2
− sin(5x) cos(−2x)
=
sin(5x) sin(−2x)
− cos(−2x)
=
= − cot(−2x)
sin(−2x)
= −(− cot(2x)) = cot(2x)
2 sin
2. Sum of Angle Identities
Use the sum of angle identities to find the exact values of the following:
(a) sin(7π/12)
First 7π/12 = 3π/12 + 4π/12 = π/4 + π/3 so use the sum of angle identity
sin(a + b) = sin(a) cos(b) + sin(b) cos(a)
to get that
sin(7π/12) = sin(π/4 + π/3)
= sin(π/4) cos(π/3) + sin(π/3) cos(π/4)
√ ! √ ! √ !
2
1
3
2
=
+
2
2
2
2
√
√
2+ 6
=
4
(b) tan(15◦ )
First 15◦ = 45◦ − 30◦ so use the difference of angles identity
tan(a − b) =
tan(a) − tan(b)
1 + tan(a) tan(b)
to get that
√
√
tan(45◦ ) − tan(30◦ )
1 − 3/3
3− 3
√
√
tan(15 ) = tan(45 − 30 ) =
=
=
1 + tan(45◦ ) tan(30◦ )
1 + (1)( 3/3)
3+ 3
√
√
√
√
(3 − 3)2
9−6 3+3
12 − 6 3
√
√ =
=
=
=2− 3
9
−
3
6
(3 + 3)(3 − 3)
◦
◦
◦
2
Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz
3
3. Triple Angle Identities
Prove the following triple angle identity:
sin(3x) = 3 sin(x) − 4 sin3 (x)
To prove this identity first use the sum of angle formulas to get
sin(3x) = sin(2x + x) = sin(2x) cos(x) + sin(x) cos(2x)
Next use the double angle identities sin(2x) = 2 sin(x) cos(x) and cos(2x) = 1 − 2 sin2 (x) to get
sin(3x) = sin(2x) cos(x) + sin(x) cos(2x)
= [2 sin(x) cos(x)] cos(x) + sin(x)[1 − 2 sin2 (x)]
= 2 sin(x) cos2 (x) + sin(x) − 2 sin3 (x)
= 2 sin(x)(1 − sin2 (x)) + sin(x) − 2 sin3 (x)
= 2 sin(x) − 2 sin3 (x) + sin(x) − 2 sin3 (x)
= 3 sin(x) − 4 sin3 (x)
4. Sums of Sine and Cosine Waves
Consider the function
√
3 sin(x) − 3 3 cos(x)
We can write√this as a single sine wave by first plotting the
point (3, −3 3) and creating a right triangle as shown.
By the Pythagorean Theorem we have that
q
√
√
r = (3)2 + (−3 3)2 = 36 = 6
And the reference angle for the triangle is π/3 so the actual
angle in the fourth quadrant is
φ=
So sin(5π/3) =
√
−3 3
6
5π
3
and cos(5π/3) = 63 .
Putting it together along with the sum of angles formula we get that
√
3 sin(x) − 3 3 cos(x) = 6
!
√
3
−3 3
sin(x) +
cos(x)
6
6
= 6(cos(5π/3) sin(x) + sin(5π/3) cos(x))
= 6 sin(x + 5π/3)
Is a sine wave with amplitude 6 and phase shift 5π/3.
(note: Some prefer to use the angle −π/3 as opposed to 5π/3, but since both angles have the same terminal point they
are equivalent.)
Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz
5. Inverse Tangent
Consider the tangent function, f (x) = tan(x) on the domain (−π/2, π/2).
The graph of f (x) is as shown:
(a) Domain: (−π/2, π/2).
(b) Range: (−∞, ∞).
(c) The graph passes the horizontal line test so it is a
one-to-one function and has and inverse.
(d) There are two vertical asymptotes at
x=
−π
2
and x =
π
2
The inverse function is
f −1 (x) = tan−1 (x)
• Domain: (−∞, ∞)
• Range: (−π/2, π/2)
• Horizontal asymptotes at
y=
−π
2
and y =
π
2
6. Inverse Trigonometric Functions
(a) Find an algebraic expression for cos(tan−1 (x)) an where x is any real number.
Let θ = tan−1 (x) so
tan(θ) = x =
x
1
Thus θ forms a right triangle as shown.
Find the hypotenuse of the right triangle from the
Pythagorean Theorem
h2 = 11 + x2
p
h = 1 + x2
Thus we have that
cos(θ) = cos(tan−1 (x)) = √
1
1 + x2
4
Math 147 - Assignment 11 Solutions - Spring 2012 - BSU - Jaimos F Skriletz
(b) Find an algebraic expression for cot(sin−1 (x)) where −1 ≤ x ≤ 1.
Let θ = sin−1 (x) so
sin(θ) = x =
x
1
Thus θ forms a right triangle as shown.
Find the adjacent side of the right triangle from the
Pythagorean Theorem
12 = a2 + x2
p
a = 1 − x2
Thus we have that
√
cot(θ) = cot(sin−1 (x)) =
1 − x2
x
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