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Chapter 8 Applications of Trigonometric Functions

Chapter 8 Applications of Trigonometric Functions. By Hannah Chung and Evan Jaques. Section 8.1 Applications Involving Right Triangles. A right triangle is a triangle with a 90 ⁰ angle . The side opposite the right angle is called the hypotenuse and the other sides are legs.

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Chapter 8 Applications of Trigonometric Functions

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  1. Chapter 8Applications of Trigonometric Functions By Hannah Chung and Evan Jaques

  2. Section 8.1 Applications Involving Right Triangles • A right triangle is a triangle with a 90⁰ angle. The side opposite the right angle is called the hypotenuse and the other sides are legs. • The angle Θ is an acute angle, or less than 90⁰. Hypotenuse Leg Leg

  3. Section 8.1 Applications Involving Right Triangles The 6 Trigonometric Functions Hypotenuse Opposite Θ Adjacent Sin Θ=Opposite -------------------------------------------------- Hypotenuse Csc=Hypotenuse ------------------------------------------------------------------- Opposite Cos Θ=Adjacent ------------------------------------------------- Hypotenuse Sec Θ=Hypotenuse ------------------------------------------------------------------- Adjacent Tan Θ=Opposite -------------------------------------------------- Adjacent Cot Θ=Adjacent ------------------------------------------------- Opposite

  4. Section 8.1 Applications Involving Right Triangles • The sum of two complementary angles is a right angle. • Cofunctions are the functions sine and cosine, tangent and cotangent, and secant and cosecant. • Cofunctions of complementary angles are equal.

  5. Section 8.2 The Law of Sines • Oblique trianglesare triangles without a right angle. • In order to solve an oblique triangle, you need to know the length of 1 side and: 2 angles, 1 angle and 1 more side, or 2 more sides. • The Law of Sines is: sin A = sin B = sin C • ---------------------------------------------- ----------------------------------------------- ----------------------------------------------- • a b c

  6. Section 8.2 The Law of Sines 60⁰ Angle C found by 180⁰=40⁰+60⁰+C Angle C=80⁰ 4 c Side b found by sin 40⁰=sin 60⁰ -------------------------------------- -------------------------------------- 4 b b=4 sin 60° -------------------------------------------------------- sin 40° C 40⁰ b Side c found by sin 40⁰=sin 80⁰ ------------------------------------- ------------------------------------- 4 c c=4 sin 80° --------------------------------------------------------- sin 40°

  7. Section 8.2 The Law of Sines Angle C found by 35°+15°+C=180° C=130° 15⁰ 5 a Side a found by sin 35° = sin 130° --------------------------------------------- ---------------------------------------------- a 5 a=5 sin 35° ------------------------------------------------------- sin 130° 35⁰ C b Side b found by sin 15° = sin 130° ---------------------------------------------- --------------------------------------------- b 5 b=5 sin 15° ------------------------------------------------------ sin 130°

  8. Section 8.2 The Law of Sines SSA triangles can be solved by using the relationship between side a, side b, and height b b a a A A a<h h=b sin A Results in no triangle a=h=b sin A Results in right triangle b b a a a A A a>b --------- Results in one triangle b sin A < a and a<b Results in 2 different triangles

  9. Section 8.2 The Law of Sines Angle C found by 25.4°+40°+C=180° C=114.6° Angle B found by sin 40° = sin B ------------------------------------------ ------------------------------ 3 2 Sin B = 2 sin 40° =.43 -------------------------------------------------------------- 3 B=25.4° Side c found by sin 40° = sin 114.6° ----------------------------------------------- -------------------------------------------------------- 3 c c=3 sin 114.6° = 4.24 ------------------------------------------------------------------------- sin 40° 3 C 2 40° B c

  10. Section 8.3 The Law of Cosines Derive the law of cosines to solve oblique triangles with two sides and the included angle (SAS) and where three sides are known (SSS) Theorem: Law of Cosines For a triangle with sides a, b, c, and opposite angles A, B, C, respectively c²= a²+ b² - 2ab cos C b²= a²+ c² - 2ac cos B a²= b²+ c² - 2bc cos A Theorem: Law of Cosines The square of one side of a triangle equals the sum of the squares of the other two sides minus twice their product times the cosine of their included angle.

  11. Section 8.3 The Law of Cosines Proof: Use distance formula to compute c². y (a cosC, a sinC) C² = (b- a cos C)² + (0 – a sinC)² = b² - 2ab cos C + a²cos²C + a²sin²C = b² - 2ab cosC + a²(cos²C + sin²C) = a² + b² - 2ab cos C a c C x (0,0) b (b, 0)

  12. Section 8.3 The Law of Cosines Use Law of Cosines to solve a SAS Triangle B Solve triangle with a=2, b=3, C=60⁰ c Law of cosines to find third side, c. c²= a²+ b² - 2ab cosC = 4+ 9 – 2 x 2 x 3 x cos60⁰ = 13-(12 x ½) = 7 c= √7 2 60⁰ A 3 For A: a²= b²+ c² - 2bc cos A 2bc cos A= b²+ c²- a² cos A= (b²+ c²- a²)/(2bc)= (9+ 7- 4)/(2x 3√7)= 12/(6√7)= (2√7)/7 A= cos¯¹(2√7/7)≈40.9⁰ For B: b²= a²+ c² - 2ac cos B cosB= (a²+ c² - b²)/2ac= (4+7- 9)/4√7= 1/(2√7)= √7/14 B= cos¯¹(√7/14)≈ 79.1⁰ Check: A+ B+ C= 40.9⁰ + 79.1⁰ + 60⁰= 180⁰

  13. Section 8.3 The Law of Cosines Use Law of Cosines to solve a SSS Triangle B For A: cosA= (b²+ c²- a²)/2bc= (9+36-16)/(2x3x6)= 29/36 A= cos¯¹29/36≈ 36.3⁰ 6 4 For B: cosB= a²+c²-b²/2ac= 16+36-9/2x4x6= 43/48 B= cos¯¹43/48≈ 26.4⁰ C A 3 Since A+B+C= 180⁰, C=180⁰ - A – B= 180⁰- 36.3⁰-26.4⁰= 117.3⁰

  14. Section 8.4 Area of a Triangle The area K of a triangle is: K= 1/2bh where b is the base and h is an altitude drawn to that base h b Proof: the area of the triangle with base b and altitude h is exactly half the area of the rectangle, which is bh.

  15. Section 8.4 Area of a Triangle Find the area of SAS triangles Suppose that we know two sides a and b and the angle C. Then the altitude can be found through: h/a= sinC So that h= asinC Using this fact in formula (1) produces K=1/2bh= 1/2b(asinC)= 1/2absinC So: K=1/2absinC (2) By dropping altitudes from the other two vertices of the triangle, we obtain: K=1/2bcsinA K=1/2acsinB a h C b Theorem: The area K of a triangle equals one-half the product of two of its sides times the sine of their included angle.

  16. Section 8.4 Area of a Triangle Find the area of SSS triangles If three sides of a triangle are known, we use Heron’s Formula: The area K of a triangle with sides a, b, and c is K= √s(s-a)(s-b)(s-c) Where s=1/2(a+b+c) Example: a b a=4, b=5, c=7 s=1/2(4+5+7)= 8 K= √s(s-a)(s-b)(s-c)= K=√8x4x3x1= √96= 4√6 units c

  17. Section 8.5 Simple Harmonic Motion; Damped Motion; Combining Waves • Simple harmonic motion is a special kind of vibrational motion in which the acceleration a of the object is directly proportionate to the negative of its displacement d from its rest position. a=-kd • An object that moves on a coordinate axis so that the distance d from its rest position at time t is given by either d=a cos (ωt) or d = a sin (ωt), where a and ω>0 are constants. • The motion has amplitude |a| and period 2π/ω. • The frequency f of an object in simple harmonic motion is the number of oscillations per unit time. Since the period is the time required for one oscillation, it follows that the frequency is the reciprocal of the period. • f=ω/2π

  18. Section 8.5 Simple Harmonic Motion; Damped Motion; Combining Waves • Suppose that an object attached to a coiled spring is pulled down a distance of 5 inches from its rest position and then released. If the time for one oscillation is 3 seconds, write an equation that relates the displacement d of the object from its rest position after time t. Assume no friction. • The motion is simple harmonic. When the object is released (t=0), the displacement of the object from the rest position is -5 units (since the object was pulled down). Because d=-5 when t=0, it is easier to use cosine d = a cos (ωt) to describe the motion. The amplitude is |-5|=5 and the period is 3 so a=-5 and 2π/ω=period=3, ω=2π/3. • The equation of the motion of the object is d=-5 cos [2π t] • --------------------------------------------------- • 3

  19. Section 8.5 Simple Harmonic Motion; Damped Motion; Combining Waves • Damped motionis simple harmonic motion with friction. • The displacement d of an oscillating object from its at-rest position at time t is given by • b is the damping factor or damping coefficient and m is the mass of the oscillating object. Here |a| is the displacement at t=0 and 2π/ω is the period under simple harmonic motion (no damping). ω^2 – b^2 t ------------------------- 4m^2

  20. Section 8.5 Simple Harmonic Motion; Damped Motion; Combining Waves • Analyze the damped vibration curve d(t)=e^(-t/π) cos t, t ≥ 0 • The displacement d is the product of y=e^(-t/π) and y=cos t. Using absolute value properties and the fact that |cos t| ≤ 1, we find that • |d(t)|=|e^(-t/π) cos t|=|e^(-t/π||cos t|≤|e^(-t/π|=e^(-t/π) • As a result, -e^(-t/π)≤d(t)≤e^(-t/π) • This means that the graph of d will lie between the graphs of y=e^(-t/π) and y=-e^(-t/π), which are the bounding curves of d.

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