Exercise 66 Page 885

We want to find the exact value of csc θ given that cos θ =- 35. To do so, we will use one of the Pythagorean Identities and one of the Reciprocal Identities. cos ^2 θ +sin ^2 θ & =1 sin θ & = 1/csc θ, csc θ ≠ 0 Let's do it!

cos θ =- 3/5

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LHS^2=RHS^2

RaiseEqn

LHS^2=RHS^2

cos ^2 θ = ( - 3/5 )^2

NegBaseToPosBase

(- a)^2=a^2

cos ^2 θ = ( 3/5 )^2

PowQuot

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

cos ^2 θ =9/25

AddEqn

LHS+sin ^2 θ=RHS+sin ^2 θ

cos ^2 θ + sin ^2 θ =9/25 + sin ^2 θ

Substitute

cos ^2 θ + sin ^2 θ= 1

1 =9/25 + sin ^2 θ

SubEqn

LHS-9/25=RHS-9/25

1 -9/25 = sin ^2 θ

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Subtract term

OneToFrac

Rewrite 1 as 25/25

25/25 -9/25 = sin ^2 θ

SubFrac

Subtract fractions

16/25 = sin ^2 θ

Substitute

sin θ= 1/csc θ

16/25 = ( 1/csc θ ) ^2

PowQuot

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

16/25 = 1/csc ^2 θ

MultEqn

LHS * csc ^2 θ=RHS* csc ^2 θ

16/25 * csc ^2 θ =1

MultEqn

LHS * 25/16=RHS* 25/16

16/25 * csc ^2 θ * 25/16=25/16

MultFracByInverse

a/b* b/a=1

csc ^2 θ=25/16

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sqrt(LHS)=sqrt(RHS)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

csc θ =± sqrt(25/16)

SqrtQuot

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

csc θ =± sqrt(25)/sqrt(16)

CalcRoot

Calculate root

csc θ =± 5/4

Be aware that we are told that θ lies between 90^(∘) and 180^(∘). Therefore, θ is in Quadrant II.

In this quadrant, the sine of θ is positive. Since csc θ= 1sin θ, the sign of csc θ is also positive in this quadrant. Therefore, we will only keep the positive solution. csc θ =5/4