Sums and differences as products (deg)

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.

$\sin\left(A+B\right)\equiv\sin A\cos B+\sin B\cos A$sin(A+B)≡sinAcosB+sinBcosA	$\left(1\right)$(1)
$\sin\left(A-B\right)\equiv\sin A\cos B-\sin B\cos A$sin(A−B)≡sinAcosB−sinBcosA	$\left(2\right)$(2)
$\cos\left(A+B\right)\equiv\cos A\cos B-\sin A\sin B$cos(A+B)≡cosAcosB−sinAsinB	$\left(3\right)$(3)
$\cos\left(A-B\right)\equiv\cos A\cos B+\sin A\sin B$cos(A−B)≡cosAcosB+sinAsinB	$\left(4\right)$(4)

 

 

 

 

If we add (1) and (2), we have 

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

$\left(5\right)$(5)

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

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

$\left(6\right)$(6)

 

 

and by subtraction,

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

$\left(7\right)$(7)

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

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

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

 

 

 

 

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

$\sin U-\sin V=2\sin\frac{U-V}{2}\cos\frac{U+V}{2}$sinU−sinV=2sinU−V2​cosU+V2​

$\left(11\right)$(11)

 

 

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

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

On comparing this with the original expression, we see that $a=r\cos\alpha$a=rcosα and $b=r\sin\alpha$b=rsinα.

Hence, $r=\sqrt{a^2+b^2}$r=√a2+b2 and $\tan\alpha=\frac{b}{a}$tanα=ba​. Then, using the notation $\tan^{-1}$tan−1 for the inverse tangent function, we can write

$$	$\left(12\right)$(12)

 

 

Example

Express $\cos255^\circ-\cos45^\circ$cos255°−cos45° more simply.

Using (10), $\cos V-\cos U=2\sin\frac{U+V}{2}\sin\frac{U-V}{2}$cosV−cosU=2sinU+V2​sinU−V2​, we have 

$\cos255^\circ-\cos45^\circ=2\sin\frac{255^\circ+45^\circ}{2}\sin\frac{45^\circ-255^\circ}{2}$cos255°−cos45°=2sin255°+45°2​sin45°−255°2​

That is,

$\cos255^\circ-\cos45^\circ=$cos255°−cos45°=	$2\sin150^\circ\sin\left(-105\right)^\circ$2sin150°sin(−105)°
$=$=	$-2\sin30^\circ\sin75^\circ$−2sin30°sin75°
$=$=	$-\sin75^\circ$−sin75°

Using a half-angle formula, $$ we can further simplify this to the exact value $-\frac{1}{2}\sqrt{2+\sqrt{3}}$−12​√2+√3.

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3