New Zealand
Level 8 - NCEA Level 3

# Sums and differences as products

Lesson

### Products

If the sine and cosine sum and difference formulas are written down side-by-side it becomes apparent that useful results can be obtained by adding some of them them in pairs.

 $\sin\left(A+B\right)$sin(A+B) $=$= $\sin\left(A\right)\cos\left(B\right)+\sin\left(B\right)\cos\left(A\right)$sin(A)cos(B)+sin(B)cos(A) (1) $\sin\left(A-B\right)$sin(A−B) $=$= $\sin\left(A\right)\cos\left(B\right)-\sin\left(B\right)\cos\left(A\right)$sin(A)cos(B)−sin(B)cos(A) (2) $\cos\left(A+B\right)$cos(A+B) $=$= $\cos\left(A\right)\cos\left(B\right)-\sin\left(A\right)\sin\left(B\right)$cos(A)cos(B)−sin(A)sin(B) (3) $\cos\left(A-B\right)$cos(A−B) $=$= $\cos\left(A\right)\cos\left(B\right)+\sin\left(A\right)\sin\left(B\right)$cos(A)cos(B)+sin(A)sin(B) (4)

If we add (1) and (2), we have

 $\sin\left(A+B\right)+\sin\left(A-B\right)$sin(A+B)+sin(A−B) $=$= $2\sin\left(A\right)\cos\left(B\right)$2sin(A)cos(B) (5)

Similarly, from (3) and (4) we obtain, by addition,

 $\cos\left(A+B\right)+\cos\left(A-B\right)$cos(A+B)+cos(A−B) $=$= $2\cos\left(A\right)\cos\left(B\right)$2cos(A)cos(B) (6)

and by subtraction,

 $\cos\left(A-B\right)-\cos\left(A+B\right)$cos(A−B)−cos(A+B) $=$= $2\sin\left(A\right)\sin\left(B\right)$2sin(A)sin(B) (7)

Equations (5), (6) and (7) give the following three product formulas:

 $\sin\left(A\right)\cos\left(B\right)$sin(A)cos(B) $=$= $\frac{1}{2}\left(\sin\left(A+B\right)+\sin\left(A-B\right)\right)$12​(sin(A+B)+sin(A−B)) (5a) $\cos\left(A\right)\cos\left(B\right)$cos(A)cos(B) $=$= $\frac{1}{2}\left(\cos\left(A+B\right)+\cos\left(A-B\right)\right)$12​(cos(A+B)+cos(A−B)) (6a) $\sin\left(A\right)\sin\left(B\right)$sin(A)sin(B) $=$= $\frac{1}{2}\left(\cos\left(A-B\right)-\cos\left(A+B\right)\right)$12​(cos(A−B)−cos(A+B)) (7a)

### Sums

By re-writing (5a), (6a) and (7a) we can obtain formulas for the sums and differences of sines and cosines. To do this, we let $U=A+B$U=A+B and $V=A-B$V=AB. Then, by solving these equations for $A$A and $B$B we get $A=\frac{U+V}{2}$A=U+V2 and $B=\frac{U-V}{2}$B=UV2.

Thus, by substituting for $A$A and $B$B in the product formulas and rearranging slightly, we obtain:

 $\sin\left(U\right)+\sin\left(V\right)$sin(U)+sin(V) $=$= $2\sin\left(\frac{U+V}{2}\right)\cos\left(\frac{U-V}{2}\right)$2sin(U+V2​)cos(U−V2​) (8) $\cos\left(U\right)+\cos\left(V\right)$cos(U)+cos(V) $=$= $2\cos\left(\frac{U+V}{2}\right)\cos\left(\frac{U-V}{2}\right)$2cos(U+V2​)cos(U−V2​) (9) $\cos\left(V\right)-\cos\left(U\right)$cos(V)−cos(U) $=$= $2\sin\left(\frac{U+V}{2}\right)\sin\left(\frac{U-V}{2}\right)$2sin(U+V2​)sin(U−V2​) (10)

and from (8), using the fact that $-\sin\left(V\right)=\sin\left(-V\right)$sin(V)=sin(V), we can write

 $\sin\left(U\right)-\sin\left(V\right)$sin(U)−sin(V) $=$= $2\sin\left(\frac{U-V}{2}\right)\cos\left(\frac{U+V}{2}\right)$2sin(U−V2​)cos(U+V2​) (11)

Another type of sum, with a very useful simplification, occurs between different multiples of the sine and cosine of identical angles.

The expression $a\sin\left(\theta\right)+b\cos\left(\theta\right)$asin(θ)+bcos(θ) can be written in the form $r\sin\left(\theta+\alpha\right)$rsin(θ+α). The latter expands to $r\left(\sin\left(\theta\right)\cos\left(\alpha\right)+\cos\left(\theta\right)\sin\left(\alpha\right)\right)$r(sin(θ)cos(α)+cos(θ)sin(α)).

On comparing this with the original expression, we see that $a=r\cos\left(\alpha\right)$a=rcos(α) and $b=r\sin\left(\alpha\right)$b=rsin(α).

Hence, $r=\sqrt{a^2+b^2}$r=a2+b2 and $\tan\left(\alpha\right)=\frac{b}{a}$tan(α)=ba. Then, using the notation $\tan^{-1}$tan1 for the inverse tangent function, we can write

 $a\sin\left(\theta\right)+b\cos\left(\theta\right)$asin(θ)+bcos(θ) $=$= $\sqrt{a^2+b^2}\sin\left(\theta+\tan^{-1}\left(\frac{b}{a}\right)\right)$√a2+b2sin(θ+tan−1(ba​)) (12)

#### Worked example

Express $\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)$cos(17π12)cos(π4) more simply.

Do: Using (10), $\cos\left(V\right)-\cos\left(U\right)=2\sin\left(\frac{U+V}{2}\right)\sin\left(\frac{U-V}{2}\right)$cos(V)cos(U)=2sin(U+V2)sin(UV2), we have

$\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)=2\sin\left(\frac{\frac{17\pi}{12}+\frac{\pi}{4}}{2}\right)\sin\left(\frac{\frac{\pi}{4}-\frac{17\pi}{12}}{2}\right)$cos(17π12)cos(π4)=2sin(17π12+π42)sin(π417π122)

That is,

 $\cos\left(\frac{17\pi}{12}\right)-\cos\left(\frac{\pi}{4}\right)$cos(17π12​)−cos(π4​) $=$= $2\sin\left(\frac{5\pi}{6}\right)\sin\left(-\frac{7\pi}{12}\right)$2sin(5π6​)sin(−7π12​) $=$= $-2\sin\left(\frac{\pi}{6}\right)\sin\left(\frac{5\pi}{12}\right)$−2sin(π6​)sin(5π12​) $=$= $-\sin\left(\frac{5\pi}{12}\right)$−sin(5π12​)

Using a half-angle formula, $\sin\left(\frac{\theta}{2}\right)=\sqrt{\frac{1}{2}\left(1-\cos\left(\theta\right)\right)}$sin(θ2)=12(1cos(θ)) we can further simplify this to the exact value $-\frac{1}{2}\sqrt{2+\sqrt{3}}$122+3.

#### Practice questions

##### QUESTION 1

Express $\cos\left(3x+2y\right)\cos\left(x-y\right)$cos(3x+2y)cos(xy) as a sum or difference of two trigonometric functions.

##### QUESTION 2

Express $\sin\left(6x\right)+\sin\left(4x\right)$sin(6x)+sin(4x) as a product of two trigonometric functions.

##### QUESTION 3

By expressing the left-hand side of the equation as a product, solve the equation $\sin5x+\sin x=0$sin5x+sinx=0 for $0\le x$0x$<=$<=$2\pi$2π.

### Outcomes

#### M8-6

Manipulate trigonometric expressions

#### 91575

Apply trigonometric methods in solving problems