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McGraw Hill Glencoe Algebra 2, 2012

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McGraw Hill Glencoe Algebra 2, 2012 View details

1. Trigonometric Identities

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Exercise 8 Page 876

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a We are given a formula that lets us determine the intensity of the emerging light. Here, I_0 is the intensity of the light incoming to the second polarized lens, I is the intensity of the emerging light, and θ is the angle between the axes of polarization.

I=I_0-I_0/csc^2θ We want to simplify this formula in terms of cosθ. We will start by using one of the Reciprocal Identities. We will transform it so that it will better match our formula. 1/cscθ=sinθ, for cscθ≠0 [0.5em] ⇕ 1/csc^2θ=sin^2θ, for cscθ≠0 Let's use this identity to transform our formula!

I=I_0-I_0/csc^2θ

a/b=a* 1/b

I=I_0-I_0 * 1/csc^2θ

1/csc^2θ= sin^2θ

I=I_0-I_0 sin^2θ

We have obtained a formula in terms of sinθ. Recall the Pythagorean Identity considering both sin^2θ and cos^2θ. We can isolate sin^2θ from it and substitute the obtained expression into our formula. cos^2θ+sin^2θ=1 ⇕ sin^2θ= 1-cos^2θ Let's substitute 1-cos^2θ for sin^2θ and simplify.

I=I_0-I_0 sin^2θ

sin^2θ= 1-cos^2θ

I=I_0-I_0( 1-cos^2θ)

▼

Simplify

Distribute I_0

I=I_0-I_0+I_0cos^2θ

Subtract term

I=I_0cos^2θ

We have now obtained the formula in terms of cosθ. Note that this formula is true when cscθ≠0.

b Now we will use the simplified formula to determine the intensity of light that passes through a second polarizing lens with an axis at 30^(∘) to the original. To do so we need to substitute 30^(∘) for θ in the simplified formula and evaluate it.

I=I_0cos^2θ

θ= 30^(∘)

I=I_0cos^2 30^(∘)

▼

Evaluate right-hand side

cos^2(θ)=(cos(θ))^2

I=I_0(cos30^(∘))^2

\ifnumequal{30}{0}{\cos\left(0^\circ\right)=1}{}\ifnumequal{30}{30}{\cos\left(30^\circ\right)=\dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{45}{\cos\left(45^\circ\right)=\dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{60}{\cos\left(60^\circ\right)=\dfrac{1}{2}}{}\ifnumequal{30}{90}{\cos\left(90^\circ\right)=0}{}\ifnumequal{30}{120}{\cos\left(120^\circ\right)=\text{-} \dfrac{1}{2}}{}\ifnumequal{30}{135}{\cos\left(135^\circ\right)=\text{-} \dfrac{\sqrt{2}}{2}}{}\ifnumequal{30}{150}{\cos\left(150^\circ\right)=\text{-} \dfrac{\sqrt{3}}{2}}{}\ifnumequal{30}{180}{\cos\left(180^\circ\right)=\text{-} 1}{}\ifnumequal{30}{210}{\cos\left(210^\circ\right)=\text{-} \dfrac{\sqrt 3}2}{}\ifnumequal{30}{225}{\cos\left(225^\circ\right)=\text{-} \dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{240}{\cos\left(240^\circ\right)=\text{-} \dfrac {1}2}{}\ifnumequal{30}{270}{\cos\left(270^\circ\right)=0}{}\ifnumequal{30}{300}{\cos\left(300^\circ\right)=\dfrac{1}2}{}\ifnumequal{30}{315}{\cos\left(315^\circ\right)=\dfrac {\sqrt{2}} {2}}{}\ifnumequal{30}{330}{\cos\left(330^\circ\right)=\dfrac{\sqrt 3}2}{}\ifnumequal{30}{360}{\cos\left(360^\circ\right)=1}{}

I=I_0(sqrt(3)/2)^2

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

I=I_0* (sqrt(3))^2/4

( sqrt(a) )^2 = a

I=I_0 * 3/4

Multiply

I=3/4I_0

Subchapter links

Check Your Understanding p.876

H.O.T. Problems p.878

Standardized Test Practice p.879

Spiral Review p.879

Skills Review p.879