Sums and differences as products (rad)

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=A−B. 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=U−V2​.

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 $\tan^{-1}$tan−1 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(U−V2​), 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​+π4​2​)sin(π4​−17π12​2​)

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​(1−cos(θ)) we can further simplify this to the exact value $-\frac{1}{2}\sqrt{2+\sqrt{3}}$−12​√2+√3.

 

Practice questions

QUESTION 1

QUESTION 2

QUESTION 3