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7.4 - Sum and Difference Identities

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<strong>7.4</strong> - <strong>Sum</strong> <strong>and</strong> <strong>Difference</strong> <strong>Identities</strong><br />

Suppose that α <strong>and</strong> β acute angles such that α+β are in quadrant I. We are<br />

interested in computing sin(α+β). From our figure we see that sin(α+β) =<br />

L+cos(α)sin(β), the height of the right triangle with angle α+β.<br />

β<br />

α<br />

We need to compute L.<br />

β<br />

α<br />

cos(α)sin(β)<br />

cos(β)<br />

1<br />

γ<br />

α<br />

γ<br />

L<br />

L<br />

sin(β)<br />

sin(β)<br />

sin(α)<br />

sin(α)


L<br />

Bysimilartrianglesweseethat<br />

Hence we have<br />

sin(α)<br />

= cos(β)<br />

1 <strong>and</strong>thereforeL = sin(α)cos(β).<br />

sin(α+β) = cos(α)sin(β)+sin(α)cos(β).<br />

<strong>Sum</strong> <strong>and</strong> <strong>Difference</strong> <strong>Identities</strong><br />

We have established the following result.<br />

Theorem 1.<br />

cos(a±b) = cos(a)cos(b)∓sin(a)sin(b)<br />

sin(a±b) = sin(a)cos(b)±cos(a)sin(b)<br />

tan(a±b) = tan(a)±tan(b)<br />

1∓tan(a)tan(b)<br />

Example 1. Findcos(α+β),giventhatα<strong>and</strong>β areinquadrantIII,cos(α) =<br />

−3 12 , <strong>and</strong> tan(β) = 5 5 .<br />

We have cos(α) = − 3<br />

5<br />

−12<br />

−3<br />

−4<br />

α<br />

5<br />

−5<br />

13<br />

5<br />

, sin(α) = −4,<br />

cos(β) = − 5 13<br />

cos(α+β) = cos(α)cos(β)−sin(α)sin(β) =<br />

β<br />

<br />

− 3<br />

5<br />

<br />

− 5<br />

13<br />

<strong>and</strong> sin(β) = −12<br />

13 ,<br />

<br />

− − 4<br />

<br />

−<br />

5<br />

12<br />

13<br />

Example 2. Use a trig formula to find the exact value for the expression<br />

tan <br />

π . 12<br />

2<br />

<br />

.


We have<br />

Example 3. Verify<br />

We have<br />

<br />

π<br />

<br />

π<br />

tan = tan<br />

12 3<br />

= tan π<br />

3<br />

=<br />

π<br />

<br />

−<br />

4 <br />

π −tan 4<br />

1+tan <br />

π π<br />

tan 3 4<br />

√<br />

3−1<br />

1+ √ 3 .<br />

sin(x+y) tan(x)+tan(y)<br />

=<br />

sin(x−y) tan(x)−tan(y) .<br />

sin(x+y) sin(x)cos(y)+sin(y)cos(x)<br />

=<br />

sin(x−y) sin(x)cos(y)−sin(y)cos(x)<br />

=<br />

=<br />

sin(x)cos(y)+sin(y)cos(x)<br />

cos(x)cos(y)<br />

sin(x)cos(y)−sin(y)cos(x)<br />

cos(x)cos(y)<br />

sin(x)<br />

cos(x)<br />

+ sin(y)<br />

cos(y)<br />

sin(x) sin(y)<br />

− cos(x) cos(y)<br />

= tan(x)+tan(y)<br />

tan(x)−tan(y) .<br />

Example 4. Show that cos(x+2π) = cos(x).<br />

By the angle addition formula we get<br />

cos(x+2π) = cos(x)cos(2π)−sin(x)sin(2π) = cos(x).<br />

Co-Function <strong>Identities</strong><br />

The co-function identities are given by<br />

<br />

π<br />

sin<br />

2 −x<br />

<br />

π<br />

= cos(x) cos<br />

2 −x<br />

<br />

π<br />

= sin(x) tan<br />

2 −x<br />

<br />

= cot(x)<br />

<br />

π<br />

csc<br />

2 −x<br />

<br />

π<br />

= sec(x) sec<br />

2 −x<br />

<br />

π<br />

= csc(x) cot<br />

2 −x<br />

<br />

= tan(x).<br />

3


Example 5. Express csc( π<br />

2<br />

−x) in term of sec.<br />

<br />

π<br />

csc<br />

2 −x<br />

<br />

= sec(x).<br />

Example 6. Express sin(3x) in terms of cos.<br />

<br />

π<br />

sin(3x) = cos<br />

2 −3x<br />

<br />

.<br />

Example 7. Find <br />

x−2<br />

cos arctan .<br />

x<br />

Now<br />

tan(θ) = x−2<br />

x<br />

implies that the hypotenuse of the triangle is x2 +(x−2) 2 .<br />

Therefore cos(θ) =<br />

An Application<br />

x−2<br />

√ x<br />

x2 +(x−2) 2 .<br />

x 2 +(x−2) 2<br />

x<br />

Example 8. Sound is a result of waves applying pressure to a person’s<br />

eardrum. Foraparticularwaveradiatingoutward, thetrigonometricfunction<br />

P(r) = a cos(πr−1000t) can be used to express pressure at a radius of r feet<br />

r<br />

from the source after t seconds. In this formula, a is the maximum sound<br />

pressure at the source, measured in pounds per square foot.<br />

Use a difference identity to simplify the expression for P(r) when r is an<br />

even integer.<br />

We get<br />

cos(rπ −1000t) = cos(rπ)cos(1000t)+sin(rπ)sin(1000t)<br />

<strong>and</strong> if r is even we have rπ = 0,2π,4π,... <strong>and</strong> therefore<br />

cos(2kπ−1000t) = cos(1000t).<br />

4<br />

θ


Extra problems<br />

1. If a person with weight W bends at the waist with a straight back, then<br />

the force F exerted by the lower back muscles may be approximated<br />

using F = 2.89W sin(θ+90 ◦ ), where θ is the angle between a person’s<br />

torso <strong>and</strong> the horizontal. Approximate the value of θ that will cause a<br />

130 pound person’s back muscles to exert a force of 200 pounds.<br />

2. Given cos(s) = − 15<br />

<strong>and</strong> cos(t) = −4,<br />

<strong>and</strong> that s <strong>and</strong> t are in quadrant<br />

17 5<br />

II, find sin(s+t),cos(s+t),tan(s+t) <strong>and</strong> the quadrant of s+t.<br />

3. Find the angle of intersection, θ, for lines l1 <strong>and</strong> l2 if their slopes are<br />

given by tan(α) <strong>and</strong> tan(β) <strong>and</strong> show that θ satisfies the following,<br />

where α <strong>and</strong> β are acute angles.<br />

Khan Academy Video Links<br />

Proof of sin(a+b)<br />

Proof of cos(a+b)<br />

tan(θ) = tan(β)−tan(α)<br />

1+tan(α)tan(β)<br />

5

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