McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
2. Verifying Trigonometric Identities
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Exercise 51 Page 884

Practice makes perfect
a We are given an equation for the height of the ball, where θ is the angle between the ground and the path of the ball, v_0 is its initial velocity in meters per second and g is the acceleration due to gravity.
h=v_0 ^2sin^2θ/2g The value of g is 9.8m/s^2. We want to find the height of the ball for different values of θ if v_0 is 47 meters per second. Let's start with substituting these values into the formula and simplifying.
h=v_0 ^2sin^2θ/2g
h=47^2sin^2θ/2( 9.8)
Simplify right-hand side
h=2209sin^2θ/2(9.8)
h=2209sin^2θ/19.6
We want to evaluate this expression for different values of θ. Let's start with θ= 30^(∘).
h=2209sin^2θ/19.6
h=2209sin^2 30^(∘)/19.6
Evaluate right-hand side

sin^2(θ)=(sin(θ))^2

h=2209(sin30^(∘))^2/19.6
h=2209( 12)^2/19.6
h=2209( 14)/19.6
h=22094/19.6
h=2209/4(19.6)
h=2209/78.4
h=28.176020...
h≈ 28.2
The height of the ball for θ=30^(∘) is approximately 28.2 meters. Let's now find the heights for other angles.
θ 2209sin^2θ/19.6 h=2209sin^2θ/19.6 Approximation
45^(∘) 2209sin^2 45^(∘)/19.6 56.352040... 56.4
60^(∘) 2209sin^2 60^(∘)/19.6 84.528061... 85.5
90^(∘) 2209sin^2 90^(∘)/19.6 112.704081... 112.7
b We will now graph the equation on our graphing calculator. We want to graph it when g= 9.8 and v_0= 47. We will use the simplified form we have obtained before.
h=( 47) ^2sin^2θ/2( 9.8) ⇕ h=2209sin^2θ/19.6 For the purpose of graphing we will change h for y and θ for x.

y=2209sin^2x/19.6 In this case we will consider the x-values between 0 and 180. Also in the previous part of exercise, we have found that for x=90, y≈ 112.7. Therefore, the maximum value for y in the window needs to be significantly bigger, let's say 150. The minimum value for y will be 0. We will now push WINDOW and resize it.

Now, let's graph the equation. Press the Y= button and type the equation in the first row. Having written the equation, push GRAPH to draw them.

c Now, we want to prove that the formula is equivalent to the given one.
v_0 ^2tan^2θ/2gsec^2θRecall the definition of secant and one of the Quotient Identities considering tangent. secθ = & 1/cosθ tanθ = & sinθ/cosθ
v_0 ^2tan^2θ/2gsec^2θ? =v_0 ^2sin^2θ/2g
v_0 ^2( sinθcosθ)^2/2g( 1cosθ)^2? =v_0 ^2sin^2θ/2g
Simplify left-hand side
v_0 ^2 ( sin^2θcos^2θ)/2g( 1cos^2θ)? =v_0 ^2sin^2θ/2g
v_0 ^2/2g* sin^2θcos^2θ/1cos^2θ? =v_0 ^2sin^2θ/2g
v_0 ^2/2g* sin^2θ/cos^2θ* cos^2θ/1? =v_0 ^2sin^2θ/2g
v_0 ^2 sin^2θcos^2θ/2gcos^2θ? =v_0 ^2sin^2θ/2g
v_0 ^2 sin^2θ/2g=v_0 ^2sin^2θ/2g ✓
Therefore, the formulas are equivalent.