Exercise 65 Page 885

We want to find the exact value of sin θ given that cos θ = 23. To do so, we will use one of the Pythagorean Identities. cos^2 θ + sin^2 θ =1 Let's do it!

cos θ = 2/3

RaiseEqn

LHS^2=RHS^2

cos^2 θ = (2/3)^2

PowQuot

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

cos^2 θ = 4/9

AddEqn

LHS+sin^2 θ=RHS+sin^2 θ

cot ^2 θ +sin^2 θ= 4/9+sin^2 θ

Substitute

cot ^2 θ +sin^2 θ= 1

1= 4/9+sin^2 θ

SubEqn

LHS-4/9=RHS-4/9

1-4/9=sin^2 θ

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Subtract term

OneToFrac

Rewrite 1 as 9/9

9/9-4/9=sin^2 θ

SubFrac

Subtract fractions

5/9=sin^2 θ

RearrangeEqn

Rearrange equation

sin^2 θ = 5/9

▼

sqrt(LHS)=sqrt(RHS)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sin θ =± sqrt(5/9)

SqrtQuot

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

sin θ =± sqrt(5)/sqrt(9)

CalcRoot

Calculate root

sin θ =± sqrt(5)/3

Be aware that we are told that θ lies between 0^(∘) and 90^(∘). Therefore, θ is in Quadrant I.

In this quadrant, the sine of θ is positive. Therefore, we will only keep the positive solution. sin θ =sqrt(5)/3