McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
4. Double-Angle and Half-Angle Identities
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Exercise 1 Page 897

sin2θ=sqrt(15)/8, cos2θ=7/8, sinθ/2=sqrt(8-2sqrt(15))/4, cosθ/2=sqrt(8+2sqrt(15))/4

Practice makes perfect

Consider the given information. sin θ =1/4; 0^(∘)< θ <90^(∘) We want to find the exact values of sin 2θ, cos 2θ, sin θ2, and cos θ2. To do so, we will use some of the Pythagorean Identities, Double-Angle Identities, and Half-Angle Identities. Let's find the desired values one at a time.

Value of sin 2θ

To find the value of sin 2θ, we will use one Pythagorean Identity and one Double-Angle Identity.

Pythagorean Identity Double-Angle Identity
cos ^2 θ + sin ^2 θ =1 sin 2θ=2sin θ cos θ

Let's start by finding the value of cos θ using the Pythagorean Identity in the above table.

cos ^2 θ +sin ^2 θ =1
cos ^2 θ +( 1/4)^2 =1
â–¼
Calculate power
cos ^2 θ +1^2/4^2= 1
cos ^2 θ +1/16=1
â–¼
Solve for cos θ
cos ^2 θ=1- 1/16
cos ^2 θ =16/16- 1/16
cos ^2 θ =15/16
cos θ =± sqrt(15/16)
cos θ =± sqrt(15)/sqrt(16)
cos θ =± sqrt(15)/4

We found that cos θ = ± sqrt(15)4. Now we need to find the sign of cos θ. Recall that we were told that θ is greater than 0^(∘) and less than 90^(∘). This means that if we draw the angle θ in standard position, the terminal side will be located in Quadrant I.

We can see that cos θ is positive for θ greater than 0^(∘) and less than 90^(∘). cos θ = sqrt(15)/4 Finally, we will substitute this value and the value of sin θ into our Double-Angle Identity.

sin 2θ=2sin θ cos θ
sin 2θ=2( 1/4) ( sqrt(15)/4)
â–¼
Evaluate right-hand side
sin 2θ=2/4 (sqrt(15)/4)
sin 2θ=2sqrt(15)/16
sin 2θ=sqrt(15)/8

Value of cos 2θ

To find the value of cos 2θ, we will use only one Double-Angle Identity. cos 2θ = 2cos ^2 θ -1 We previously found that cos θ = sqrt(15)4, so we can substitute this value into the identity and simplify.

cos 2θ = 2cos ^2 θ -1
cos 2θ = 2( sqrt(15)/4)^2 -1
â–¼
Evaluate right-hand side
cos 2θ =2* (sqrt(15))^2/4^2 -1
cos 2θ =2* 15/4^2 -1
cos 2θ =2* 15/16 -1
cos 2θ =30/16 -1
cos 2θ =30/16 -16/16
cos 2θ =14/16
cos 2θ =7/8

Value of sin θ2

To find the value of sin θ2 we will use one Half-Angle Identity. sin θ/2 = ± sqrt(1-cos θ/2) We previously found that cos θ = sqrt(15)4 so we can substitute this value in the identity and simplify.

sin θ/2 = ± sqrt(1-cos θ/2)
sin θ/2 = ± sqrt(1- sqrt(15)4/2)
â–¼
Evaluate right-hand side
sin θ/2 = ± sqrt(44- sqrt(15)4/2)
sin θ/2 = ± sqrt(4-sqrt(15)4/2)
sin θ/2 = ± sqrt(4-sqrt(15)/2* 4)
sin θ/2 = ± sqrt(4-sqrt(15))/sqrt(2* 4)
sin θ/2 = ± sqrt(4-sqrt(15))/sqrt(2)* sqrt(4)
sin θ/2 = ± sqrt(4-sqrt(15))/sqrt(2)* 2
â–¼
Rationalize the denominator
sin θ/2 = ± sqrt(4-sqrt(15))* sqrt(2)/sqrt(2)* 2* sqrt(2)
sin θ/2 = ± sqrt(4-sqrt(15))* sqrt(2)/2* 2
sin θ/2 = ± sqrt((4-sqrt(15))(2))/2* 2
sin θ/2 = ± sqrt(8-2sqrt(15))/2* 2
sin θ/2 = ± sqrt(8-2sqrt(15))/4

Now let's determine the sign of sin θ2. Knowing that θ is greater than 0^(∘) and less than 90^(∘), we can determine the range of values for θ2. 0^(∘) < θ < 90^(∘) ⇔ 0^(∘) < θ/2 < 45^(∘) Since θ2 is greater than 0^(∘) and less than 45^(∘), it is located in Quadrant I.

Therefore, sin θ2 is positive. sin θ/2= sqrt(8-2sqrt(15))/4

Value of cos θ2

To find the value of cos θ2 we will use one Half-Angle Identity. cos θ/2 = ± sqrt(1+cos θ/2) We previously found that cos θ = sqrt(15)4. Let's substitute this value into the identity and simplify.

cos θ/2 = ± sqrt(1+cos θ/2)
cos θ/2 = ± sqrt(1+ sqrt(15)4/2)
â–¼
Evaluate right-hand side
cos θ/2 = ± sqrt(44+ sqrt(15)4/2)
cos θ/2 = ± sqrt(4+sqrt(15)4/2)
cos θ/2 = ± sqrt(4+sqrt(15)/2* 4)
cos θ/2 = ± sqrt(4+sqrt(15))/sqrt(2* 4)
cos θ/2 = ± sqrt(4+sqrt(15))/sqrt(2)* sqrt(4)
cos θ/2 = ± sqrt(4+sqrt(15))/sqrt(2)* 2
â–¼
Rationalize the denominator
cos θ/2 = ± sqrt(4+sqrt(15))* sqrt(2)/sqrt(2)* 2* sqrt(2)
cos θ/2 = ± sqrt(4+sqrt(15))* sqrt(2)/2* 2
cos θ/2 = ± sqrt((4+sqrt(15))(2))/2* 2
cos θ/2 = ± sqrt(8+2sqrt(15))/2* 2
cos θ/2 = ± sqrt(8+2sqrt(15))/4

Previously, we found that θ2 is located in Quadrant I, where the cosine is positive. Therefore, cos θ2 is positive. cos θ/2= sqrt(8+2sqrt(15))/4