McGraw Hill Glencoe Algebra 2, 2012
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McGraw Hill Glencoe Algebra 2, 2012 View details
3. Sum and Difference of Angles Identities
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Exercise 7 Page 888

Practice makes perfect
a We are given two functions modeling different signals.
First signal: & y=20sin(3θ+45^(∘)) Second signal: & y=20sin(3θ+225^(∘)) First we want to find the sum of these two functions. Before we do so, let's simplify each one of them by using one of the Sum and Difference Identities describing the sine of the sum of two angles. sin(A+B)=sin Acos B + cos Asin B We will now use this identity to simplify the first function.
y=20sin(3θ+45^(∘))
y=20( sin3θcos45^(∘)+cos3θsin45^(∘))
Simplify right-hand side
y=20sin3θcos45^(∘)+20cos3θsin45^(∘)

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

y=20sin3θ(sqrt(2)/2)+20cos3θsin45^(∘)

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

y=20sin3θ(sqrt(2)/2)+20cos3θ(sqrt(2)/2)
y=20(sqrt(2)/2)sin3θ+20(sqrt(2)/2)cos3θ
y=20sqrt(2)/2sin3θ+20sqrt(2)/2cos3θ
y=10sqrt(2)sin3θ+10sqrt(2)cos3θ
Let's now use the same identity to simplify the second function.
y=20sin(3θ+45^(∘))
y=20( sin3θcos225^(∘)+cos3θsin225^(∘))
Simplify right-hand side
y=20sin3θcos225^(∘)+20cos3θsin225^(∘)

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

y=20sin3θ(-sqrt(2)/2)+20cos3θsin225^(∘)

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

y=20sin3θ(-sqrt(2)/2)+20cos3θ(-sqrt(2)/2)
y=20(-sqrt(2)/2)sin3θ+20(-sqrt(2)/2)cos3θ
y=-20sqrt(2)/2sin3θ+(-20sqrt(2)/2cos3θ)
y=-10sqrt(2)sin3θ+(-10sqrt(2)cos3θ)
y=-10sqrt(2)sin3θ-10sqrt(2)cos3θ
Now after we have simplified both functions we can add them.
(10sqrt(2)sin3θ+10sqrt(2)cos3θ) + (-10sqrt(2)sin3θ-10sqrt(2)cos3θ)
Simplify
sin3θ(10sqrt(2)-10sqrt(2))+10sqrt(2)cos3θ-10sqrt(2)cos3θ
sin3θ(10sqrt(2)-10sqrt(2))+cos3θ(10sqrt(2)-10sqrt(2))
sin3θ(0)+10sqrt(2)cos3θ(0)
0+0
0
The sum of the given functions is 0.
b Interference occurs when two waves combine to have a greater or smaller amplitude than either of the component waves. When the amplitude is greater than either of the component waves, the interference is constructive. When the amplitude is smaller, the interference is destructive. Let's see how the interference works on an animation.
Interference Animation
In our case the sum of the functions is 0, so the signals cancel each other completely. Therefore, here destructive interference is occurring because the waves combine to have smaller amplitude.