Big Ideas Math Algebra 2, 2014
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Big Ideas Math Algebra 2, 2014 View details
7. Using Trigonometric Identities
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Exercise 8 Page 517

Consider using the Pythagorean Identity 1 + tan^2 θ = sec^2 θ.

sin θ = - sqrt(65)/9, cos θ = 4/9, tan θ = - sqrt(65)/4, cot θ = - 4sqrt(65)/65, csc θ = - 9sqrt(65)/65

Practice makes perfect

We want to find the values of the other five trigonometric functions of θ given that sec θ = 94 and 3π2 < θ < 2π.

Finding tan θ

To find the value of tan θ, we will use one of the Pythagorean Identities. 1 + tan^2 θ = sec^2 θWe will substitute 94 for sec θ in the above equation and solve it for tan θ. Let's do it!
1+tan^2 θ = sec^2 θ
1+tan^2 θ =( 9/4)^2
â–Ľ
Simplify
1+tan^2 θ =9^2/4^2
1+tan^2 θ =81/16
tan^2 θ =81/16-1
tan^2 θ =81/16-16/16
tan^2 θ =81-16/16
tan^2 θ =65/16
â–Ľ
Solve for tan θ
tan θ =± sqrt(65/16)
tan θ =± sqrt(65)/sqrt(16)
tan θ =± sqrt(65)/4
Be aware that we are told that θ lies between 3π2 and 2π. Therefore, θ is in Quadrant IV.
graph

In this quadrant, the sine of θ is negative and cosine of θ is positive. Therefore, tan θ = sin θcos θ is negative. We will only keep the negative solution. tan θ = - sqrt(65)/4

Finding the Other Trigonometric Ratios

Knowing that sec θ = 94 and tan θ = - sqrt(65)4, we are allowed to find the four remaining trigonometric ratios. Remember to simplify fractions and rationalize denominators, if needed.

Function Substitute Simplify
sin θ=tan θ/sec θ sin θ=- sqrt(65)4/94 sin θ = - sqrt(65)/9
cos θ=1/sec θ cos θ=1/94 cos θ=4/9
cot θ=1/tan θ cot θ=1/- sqrt(65)4 cot θ=- 4sqrt(65)/65
csc θ=sec θ/tan θ csc θ=94/- sqrt(65)4 csc θ=- 9sqrt(65)/65

Showing Our Work

Deriving Formulas and Simplifying Fractions
In case it is not clear from where some of the formulas in the leftmost column of the table came, we want to show how these equations were derived. Let's first derive the formula for sin θ. We will use basic identities to show how the right-hand side matches the left-hand side.
\sin \theta \stackrel{?}= \dfrac{\tan \theta}{\sec \theta}
\sin \theta \stackrel{?}= \tan \theta\left(\dfrac{1}{\sec \theta}\right)

tan(θ)=sin(θ)/cos(θ)

\sin \theta \stackrel{?}= {\color{#0000FF}{\dfrac{\sin \theta}{\cos \theta}}}\left(\dfrac{1}{\sec \theta}\right)

sec(θ) = 1/cos(θ)

\sin \theta \stackrel{?}= \dfrac{\sin \theta}{\cos \theta}\left(\dfrac{1}{{\color{#0000FF}{\frac{1}{\cos \theta}}}}\right)
\sin \theta \stackrel{?}= \dfrac{\sin \theta}{\cos \theta}\left(\dfrac{\cos \theta}{1}\right)
\sin \theta \stackrel{?}= \dfrac{\sin \theta \cdot \cos \theta}{\cos \theta}
\sin \theta \stackrel{?}= \dfrac{\sin \theta \cdot \cancel{{\color{#FF0000}{\cos \theta}}}}{\cancel{{\color{#FF0000}{\cos \theta}}}}
sin θ = sin θ ✓
Having proved the identity sin θ = tan θsec θ, we can prove the identity csc θ= sec θtan θ.
\csc \theta \stackrel{?}= \dfrac{1}{\sin \theta}
\csc \theta \stackrel{?}= \dfrac{1}{{\color{#0000FF}{\frac{\tan \theta}{\sec \theta}}}}
csc θ = sec θ/tan θ ✓
Moreover, we will show that cot θ = 1tan θ.
\cot \theta \stackrel{?}= \dfrac{1}{\tan \theta}
\cot \theta \stackrel{?}= \dfrac{1}{{\color{#0000FF}{ \frac{\sin \theta}{\cos \theta} }}}
\cot \theta \stackrel{?}= \dfrac{\cos \theta}{\sin \theta}
cot θ = cot θ ✓
Now we are ready to move to simplifying the above fractions, which requires different methods for each case. Let's start with sin θ.
sin θ = - sqrt(65)4/94
â–Ľ
Simplify right-hand side
sin θ = - sqrt(65)4/94
sin θ = - sqrt(65)/9
To simplify cot θ and csc θ, we need to rationalize the denominators. Let's look at how this was done for csc θ first.
csc θ=94/- sqrt(65)4
â–Ľ
Simplify right-hand side
csc θ=9/- sqrt(65)
csc θ=- 9/sqrt(65)
csc θ=- 9 sqrt(65)/sqrt(65)* sqrt(65)
csc θ=- 9 sqrt(65)/(sqrt(65))^2
csc θ=- 9 sqrt(65)/65
Let's now follow a similar procedure to rationalize the denominator of cot θ.
cot θ = 1/- sqrt(65)4
â–Ľ
Simplify right-hand side
cot θ = 4/- sqrt(65)4 * 4
cot θ = 4/- sqrt(65)
cot θ = - 4/sqrt(65)
cot θ = - 4sqrt(65)/sqrt(65) * sqrt(65)
cot θ = - 4 sqrt(65)/(sqrt(65))^2
cot θ = - 4 sqrt(65)/65