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McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
Study Guide and Review
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Exercise 10 Page 909

Consider using the Pythagorean Identity cos ^2 θ + sin ^2 θ= 1.

sin θ =- sqrt(2)/2

Practice makes perfect

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

cos ^2 θ +sin ^2 θ =1
Substitute

cos θ= sqrt(2)/2

( sqrt(2)/2)^2 +sin ^2 θ =1
â–¼
Solve for sin θ
PowQuot

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

(sqrt(2))^2/4 +sin ^2 θ =1
PowSqrt

( sqrt(a) )^2 = a

2/4 +sin ^2 θ =1
ReduceFrac

a/b=.a /2./.b /2.

1/2 +sin ^2 θ =1
SubEqn

LHS-1/2=RHS-1/2

sin ^2 θ =1-1/2
OneToFrac

Rewrite 1 as 2/2

sin ^2 θ =2/2-1/2
SubFrac

Subtract fractions

sin ^2 θ =1/2
SqrtEqn

sqrt(LHS)=sqrt(RHS)

sqrt(sin ^2 θ) =sqrt(1/2)

sqrt(a^2)=± a

sin θ =± sqrt(1/2)
SqrtQuot

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

sin θ =± 1/sqrt(2)
â–¼
Simplify right-hand side
ExpandFrac

a/b=a * sqrt(2)/b * sqrt(2)

sin θ =± 1* sqrt(2)/sqrt(2)* sqrt(2)
IdPropMult

Identity Property of Multiplication

sin θ =± sqrt(2)/sqrt(2)* sqrt(2)
MultSqrt

sqrt(a)* sqrt(a)= a

sin θ =± sqrt(2)/2

Be aware that we are told that θ lies between 270^(∘) and 360^(∘). Therefore, θ is in Quadrant IV.

In this quadrant, the sine of θ is negative. Therefore, we will only keep the negative solution. sin θ =- sqrt(2)/2

Subchapter links expand_more
arrow_right Exercises p.908-910
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Exercises 11 (Page 909)
Exercises 12 (Page 909)
Exercises 13 (Page 909)
Exercises 14 (Page 909)
Exercises 15 (Page 909)
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Exercises 33 (Page 910)
Exercises 34 (Page 910)
Exercises 35 (Page 910)
Exercises 36 (Page 910)
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Exercises 39 (Page 910)
Exercises 40 (Page 910)
Exercises 41 (Page 910)