Trig fun

Since $\cos(x) = \sin(90^{\circ} - x)$ we can rewrite the sum as

$\cos(1^{\circ})\cos(2^{\circ}) + \cos(2^{\circ})\cos(3^{\circ}) + ..... + \cos(44^{\circ})\cos(45^{\circ}) + \sin(45^{\circ})\sin(44^{\circ}) + ..... + \sin(3^{\circ})\sin(2^{\circ}) + \sin(2^{\circ})\sin(1^{\circ})$

$=\displaystyle\sum_{k=1}^{44} (\cos(k^{\circ})\cos((k + 1)^{\circ}) + \sin(k^{\circ})\sin((k + 1)^{\circ}))$

$=\displaystyle\sum_{k=1}^{44} \cos(1^{\circ}) = 44\cos(1^{\circ}) = \boxed{43.99}$ to $2$ decimal places.

After referring to @brian charlesworth sol.

To those who don't know the property :

2cosAcosB = cos(A+B)+cos(A-B)

2sinAsinB = -cos(A+B)+cos(A-B)

adding the two .... we get 2 cos(A-B)

2 (cos1cos2+ ...... cos88cos89) = 2 *44 * cos 1 ( 0.99...)

(cos1cos2+ ...... cos88cos89) = cos1 * 44 = $\boxed{43.99}$

The series S = cos 1◦ cos 2◦ + cos 2◦ cos 3◦ + · · · + cos 88◦ cos 89◦ is the same as S = sin 89◦ sin 88◦ + sin 88◦ sin 87◦ + · · · + sin 2◦ sin 1◦ Add the two, and combine elements of the two series into 88 pairs: cos α cos(α + 1◦ ) + sin α sin(α + 1◦ ) = cos 1◦ So we have 2S = 88 cos 1◦ ⇒ S = 44 cos 1◦

Knowing that cos(a)=sin(90-a) we can see that this problem can be rewritten: cos1cos2+cos2cos3.......+cos44cos45+sin1sin2+sin2sin3......sin44sin45. We can then see that the problem can be reordered to: cos1cos2+sin1sin2+cos2cos3+sin2sin3.....cos44cos45+sin44sin45. Using trigonometric identities: cos(a-b)=cosacosb+sinasinb In this case a-b=1 at all times so we can make the final answer 44cos1=43.99 as required

Notice that the first and the last in the summation will be: cos(x)cos(y) and cos(90-y)cos(90-x) , this will be the same pattern for all of the other pairings. Knowing that cos(x) = sin (90-x), We see that each of the pairs will be in the form: cos(x)cos(y) and sin(y)sin(x) If we commute the first side, cos(y)cos(x) then add to sin(y)sin(x), This will be in the form cos(y-x). knowing that y-x =1, we see that each pair will become cos 1. The only thing which remains is to count the pairs, 88/2 = 44 Therefore the whole summation = 44 cos 1 = 43.99

Solve: Cos1° Cos2°+ Cos2° Cos3°+…………+Cos88° Cos89°= Cos1° Cos2°+ Cos2° Cos3°+………+ Cos(90-2)° Cos(90-1)°=44×Cos1°=43.9932..