Exercise 4 Page 911

We want to find the exact value of sinθ given that cosθ = - 35. To do so, we will use one of the Pythagorean Identities. Note that we will need to rearrange the terms so that we can use it for our expression. cos ^2 θ +sin ^2 θ=1 ⇕ 1-cos^2θ = sin^2θ Let's do it!

cos θ = -3/5

RaiseEqn

LHS^2=RHS^2

cos ^2 θ = 9/25

SubEqn

LHS-1=RHS-1

cos ^2 θ - 1= 9/25 - 1

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Add terms

OneToFrac

Rewrite 1 as 25/25

cos ^2 θ - 1= 9/25 - 25/25

SubFrac

Subtract fractions

cos ^2 θ - 1= -16/25

MultEqn

LHS * (-1)=RHS* (-1)

-cos ^2 θ + 1= 16/25

RearrangeEqn

Rearrange equation

1-cos ^2 θ= 16/25

Substitute

1-cos ^2 θ= sin ^2 θ

sin ^2 θ= 16/25

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sqrt(LHS)=sqrt(RHS)

SqrtEqn

sqrt(LHS)=sqrt(RHS)

sin θ =± sqrt(16/25)

SqrtQuot

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

sin θ =± sqrt(16)/sqrt(25)

CalcRoot

Calculate root

sin θ =± 4/5

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. Therefore, we will only keep the positive solution. sin θ =4/5