Download Triple-Angle Formulas and Linear Combinations

Triple-Angle Formulas and
Linear Combinations
Bradley Hughes
Larry Ottman
Lori Jordan
Mara Landers
Andrea Hayes
Brenda Meery
Art Fortgang
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Printed: November 16, 2013
AUTHORS
Bradley Hughes
Larry Ottman
Lori Jordan
Mara Landers
Andrea Hayes
Brenda Meery
Art Fortgang
www.ck12.org
C ONCEPT
Concept 1. Triple-Angle Formulas and Linear Combinations
1
Triple-Angle Formulas and
Linear Combinations
Here you’ll learn to derive equations for formulas with triple angles using existing trig identities, as well as to
construct linear combinations of trig functions.
In other Concepts you’ve dealt with double angle formulas. This was useful for finding the value of an angle that was
double your well known value. Now consider the idea of a "triple angle formula". If someone gave you a problem
like this:
sin 135◦
Could you compute its value?
Keep reading, and at the end of this Concept you’ll know how to simplify equations such as this using the triple
angle formula.
Watch This
MEDIA
Click image to the left for more content.
Deriving a Triple Angle Formula
Guidance
Double angle formulas are great for computing the value of a trig function in certain cases. However, sometimes
different multiples than two times and angle are desired. For example, it might be desirable to have three times the
value of an angle to use as the argument of a trig function.
By combining the sum formula and the double angle formula, formulas for triple angles and more can be found.
Here, we take an equation which takes a linear combination of sine and cosine and converts it into a simpler cosine
function.
p
A cos x + B sin x = C cos(x − D), where C = A2 + B2 , cos D = CA and sin D = CB .
You can also use the TI-83 to solve trigonometric equations. It is sometimes easier than solving the equation
algebraically. Just be careful with the directions and make sure your final answer is in the form that is called for.
You calculator cannot put radians in terms of π.
Example A
Find the formula for sin 3x
Solution: Use both the double angle formula and the sum formula.
1
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sin 3x = sin(2x + x)
= sin(2x) cos x + cos(2x) sin x
= (2 sin x cos x) cos x + (cos2 x − sin2 x) sin x
= 2 sin x cos2 x + cos2 x sin x − sin3 x
= 3 sin x cos2 x − sin3 x
= 3 sin x(1 − sin2 x) − sin3 x
= 3 sin x − 4 sin3 x
Example B
Transform 3 cos 2x − 4 sin 2x into the form C cos(2x − D)
q
Solution: A = 3 and B = −4, so C = 32 + (−4)2 = 5. Therefore cos D = 35 and sin D = − 54 which makes the
reference angle is −53.1◦ or −0.927 radians. since cosine is positive and sine is negative, the angle must be a fourth
quadrant angle. D must therefore be 306.9◦ or 5.36 radians.The final answer is 3 cos 2x − 4 sin 2x = 5 cos(2x − 5.36).
Example C
Solve sin x = 2 cos x such that 0 ≤ x ≤ 2π using a graphing calculator.
Solution: In y =, graph y1 = sin x and y2 = 2 cos x.
Next, use CALC to find the intersection points of the graphs.
Vocabulary
Linear Combination: A linear combination is a set of terms that are added or subtracted from each other with a
multiplicative constant in front of each term.
Triple Angle Identity: A triple angle identity relates the a trigonometric function of three times an argument to a
set of trigonometric functions, each containing the original argument.
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Concept 1. Triple-Angle Formulas and Linear Combinations
Guided Practice
1. Transform 5 cos x − 5 sin x to the form C cos(x − D)
2. Transform −15 cos 3x − 8 sin 3x to the form C cos(x − D)
3. Derive a formula for tan 4x.
Solutions:
√
5
= √1 =
1. If 5 cos x − 5 sin x, then A = 5 and B = −5. By the Pythagorean Theorem, C = 5 2 and cos D = √
5 2
2
√
2 . So, because B is negative, D is in Quadrant IV. Therefore, D = 7π . Our final answer is 5 √2 cos x − 7π .
2
4
4
2. If −15 cos 3x − 8 sin 3x, then A = −15 and B = −8. By the Pythagorean
Theorem, C = 17. Because A and B
are both negative, D is in Quadrant III, which means D = cos−1 15
=
0.49
+ π = 3.63 rad. Our final answer is
17
17 cos 3(x − 3.63).
3.
tan 4x = tan(2x + 2x)
tan 2x + tan 2x
=
1 − tan 2x tan 2x
2 tan 2x
=
1 − tan2 2x
2 tan x
2 · 1−tan
2x
=
2
2 tan x
1 − 1−tan
2x
4 tan x
(1 − tan2 x)2 − 4 tan2 x
÷
1 − tan2 x
(1 − tan2 x)2
4 tan x
1 − 2 tan2 x + tan4 x − 4 tan2 x
=
÷
1 − tan2 x
(1 − tan2 x)2
4 tan x
(1 − tan2 x)2
=
·
1 − tan2 x 1 − 6 tan2 x + tan4 x
4 tan x − 4 tan3 x
=
1 − 6 tan2 x + tan4 x
=
Concept Problem Solution
Using the triple angle formula we learned in this Concept for the sine function, we can break the angle down into
three times a well known angle:
sin 3x = 3 sin x − 4 sin3 x
we can solve this problem.
3
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sin(3 × 45◦ ) = 3 sin 45◦ − 4 sin3 45◦
√
√ !3
2
2
=3
−4
2
2
!
√
2
4(2)2/3
=3
−
2
8
√
√
3 2−2 2
=
2 √
2
=
2
Practice
Transform each expression to the form C cos(x − D).
1.
2.
3.
4.
5.
6.
7.
3 cos x − 2 sin x
2 cos x − sin x
−4 cos x + 5 sin x
7 cos x − 6 sin x
11 cos x + 9 sin x
14 cos x + 2 sin x
−2 cos x − 4 sin x
Derive a formula for each expression.
8.
9.
10.
11.
12.
sin 4x
cos 6x
cos 4x
csc 2x
cot 2x
Find all solutions to each equation in the interval [0, 2π).
13. cos x + cos 3x = 0
14. sin 2x = cos 3x
15. cos 2x + cos 4x = 0
4