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Core Connections Integrated III, 2015

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Core Connections Integrated III, 2015 View details

1. Section 9.1

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Exercise 92 Page 461

Practice makes perfect

a From the exercise, we have been given some information which we can use to illustrate the angle. In the unit circle, cosine is given by the horizontal axis. Therefore, the angle should coincide with a vertical line through cos θ=- 1213. Let's illustrate this on the unit circle.

Using the Pythagorean Identity, we can determine the sine value. From the diagram, we know that this value should be negative.

cos^2 θ+sin^2 θ=1

cos θ= - 12/13

( - 12/13)^2 +sin^2 θ=1

▼

Solve for sin θ

(a/b)^m=a^m/b^m

144/169 +sin^2 θ=1

LHS-144/169=RHS-144/169

sin^2 θ=1-144/169

Rewrite 1 as 169/169

sin^2 θ=169/169-144/169

Subtract fractions

sin^2 θ=25/169

sqrt(LHS)=sqrt(RHS)

sinθ=± sqrt(25/169)

sqrt(a/b)=sqrt(a)/sqrt(b)

sinθ=± 5/13

sinθ < 0

sinθ=- 5/13

The value of sinθ is - 513.

b The tangent of an angle is defined as the ratio of the opposite side to the adjacent side.

tan θ = sin θ/cos θ Since we know both the cosine and sine value, we can determine the tangent value.

tan θ =sin θ/cos θ

sin θ= - 5/13, cos θ= - 12/13

tan θ=- .5 /13./- .12 /13.

- a/- b=a/b

tan θ=.5 /13./.12 /13.

a/b=a * 13/b * 13

tan θ=5/12

Subchapter links

9.1.1 Introduction to Periodic Models p.433-434

9.1.2 Graphing the Sine Function p.438-440

9.1.3 Unit Circle ↔ Graph p.442-444

9.1.4 Graphing and Interpreting the Cosine Function p.447-449

9.1.5 Defining a Radian p.452-453

9.1.6 Building a Unit Circle p.456-457

9.1.7 The Tangent Function p.460-461