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.

$$sin(A+B)≡sinAcosB+sinBcosA	$$(1)
$$sin(A−B)≡sinAcosB−sinBcosA	$$(2)
$$cos(A+B)≡cosAcosB−sinAsinB	$$(3)
$$cos(A−B)≡cosAcosB+sinAsinB	$$(4)

 

 

 

 

If we add (1) and (2), we have 

$$sin(A+B)+sin(A−B)=2sinAcosB

$$(5)

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

$$cos(A+B)+cos(A−B)=2cosAcosB

$$(6)

 

 

and by subtraction,

$$cos(A−B)−cos(A+B)=2sinAsinB

$$(7)

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

$$sinAcosB=12​(sin(A+B)+sin(A−B))	$$(5a)
$$cosAcosB=12​(cos(A+B)+cos(A−B))	$$(6a)
$$	$$(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 $$U=A+B and $$V=A−B. Then, by solving these equations for $$A and $$B we get $$A=U+V2​ and $$B=U−V2​. 

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

$$sinU+sinV=2sinU+V2​cosU−V2​	$$(8)
$$cosU+cosV=2cosU+V2​cosU−V2​	$$(9)
$$cosV−cosU=2sinU+V2​sinU−V2​	$$(10)

 

 

 

 

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

$$sinU−sinV=2sinU−V2​cosU+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 $$asinθ+bcosθ  can be written in the form $$rsin(θ+α). The latter expands to $$r(sinθcosα+cosθsinα).

On comparing this with the original expression, we see that $$a=rcosα and $$b=rsinα.

Hence, $$r=√a2+b2 and $$tanα=ba​. Then, using the notation $$tan−1 for the inverse tangent function, we can write

$$

$$(12)

 

 

Example

Express $$cos255°−cos45° more simply.

Using (10), $$cosV−cosU=2sinU+V2​sinU−V2​, we have 

$$cos255°−cos45°=2sin255°+45°2​sin45°−255°2​

That is,

$$cos255°−cos45°=	$$2sin150°sin(−105)°
$$=	$$−2sin30°sin75°
$$=	$$−sin75°

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

Worked Examples

QUESTION 1

QUESTION 2

QUESTION 3