Question

20 баллов! 1. cos(a+b)cos(a-b)+sin(a+b)*sin(a-b) 2. tg(a+b)-tgb/1+tg(a+b)tgb 3. cos(a-п/6)-0,5sina

Answer 1

1
cos(a+b)cos(a-b)+sin(a+b)*sin(a-b)=cos(a+b-a+b)=cos2b
2
(tg(a+b)-tgb)/(1+tg(a+b)tgb)=tg(a+b-b)=tga
3
cos(a-п/6)-0,5sina=cosacosπ/6+sinasinπ/6-0,5sina=
=√3/2*cosa+0,5sina-0,5sina=√3/2*cosa

Answer 2

$\mathtt{cos(a+b)cos(a-b)+sin(a+b)sin(a-b)=}\\\mathtt{cos[(a+b)-(a-b)]=cos[a+b-a+b]=cos2b}$

$\mathtt{\frac{tg(a+b)-tgb}{1+tg(a+b)tgb}=tg[a+b-b]=tga}$

$\mathtt{cos(a-\frac{\pi}{6})-\frac{1}{2}sina=cosacos(\frac{\pi}{6})+sinasin(\frac{\pi}{6})-\frac{1}{2}sina=}\\\mathtt{cosa*\frac{\sqrt{3}}{2}+sina*\frac{1}{2}-\frac{1}{2}sina=\frac{\sqrt{3}}{2}cosa}$

Answer 3

1. Используя формулу тригонометрии для cos(a+b) и cos(a-b), получаем:

cos(a+b)cos(a-b) + sin(a+b)sin(a-b) = (cos(a)cos(b) - sin(a)sin(b))(cos(a)cos(b) + sin(a)sin(b)) + (sin(a)cos(b) + cos(a)sin(b))(sin(a)cos(b) - cos(a)sin(b)) = cos^2(a)cos^2(b) - sin^2(a)sin^2(b) + sin^2(a)cos^2(b) - cos^2(a)sin^2(b) = cos^2(a)cos^2(b) + sin^2(a)cos^2(b) - cos^2(a)sin^2(b) - sin^2(a)sin^2(b) = cos^2(a)(cos^2(b) + sin^2(b)) + sin^2(a)(cos^2(b) + sin^2(b)) - sin^2(a)sin^2(b) - cos^2(a)sin^2(b) = cos^2(a) + sin^2(a) - sin^2(a)sin^2(b) - cos^2(a)sin^2(b) = 1 - sin^2(a)sin^2(b) - cos^2(a)sin^2(b) = 1 - sin^2(b)(sin^2(a) + cos^2(a)) = 1 - sin^2(b)

Таким образом, cos(a+b)cos(a-b) + sin(a+b)sin(a-b) = 1 - sin^2(b).

2. Раскрыв знаменатель и использовав формулу тангенса для суммы, получим:

tg(a+b) - tgb/(1 + tg(a+b)tgb) = (tg(a) + tg(b))/(1 - tg(a)tg(b)) - tgb/(1 + tg(a+b)tgb) = (tg(a) + tg(b) - tgbtg(a+b) - tgb)/(1 + tg(a+b)tgb - tg(a)tg(b) - tgbtg(a+b)) = (tg(a) + tg(b) - tgb(tg(a) + tg(b)))/(1 - tg(a)tg(b) - tgb^2) = ((tg(a) + tg(b))(1 - tgb))/(1 - tg(a)tg(b) - tgb^2)

3. Используя формулу косинуса для разности углов и то, что cos(п/6) = √3/2, получим:

cos(a-п/6) - 0.5sin(a) = cos(a)cos(п/6) + sin(a)sin(п/6) - 0.5sin(a) = √3/2cos(a) + 0.5sin(a) - 0.5sin(a) = √3/2cos(a)