Question

Cos a=12/13, -п/2<а<0; Найти:sin2a;ctg2a​

Answer 1

Ответ:

Объяснение:

а)sinα/2=±√(1-cosα)/2,   α/2∈1 четв. sinα/2=+√(1+12/13)/2=

√25/26=5/√26.

б)tg2α=2tgα/(1-tg²α),  ctg2α=1/tg2α

tgα=sinα/cosα=+√1-144/ 169  :(-12/13)=5/13: (-12/13)= - 5/12.

tg2α=2·(-5/12):(1-25/144)= -5/6: 119/144= -120/119.

ctg2α=- 119/120.

Answer 2

To find the values of sin(2a) and ctg(2a), we'll use the given information that a = 12/13 and -π/2 < a < 0.

First, let's find sin(2a):

sin(2a) = 2 * sin(a) * cos(a)

We already know cos(a) = 12/13, so we need to find sin(a). To do that, we'll use the Pythagorean identity:

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

sin^2(a) + (12/13)^2 = 1

sin^2(a) = 1 - (12/13)^2

sin^2(a) = 1 - 144/169

sin^2(a) = (169 - 144)/169

sin^2(a) = 25/169

sin(a) = sqrt(25/169) = 5/13 (since sin(a) is positive in the given range)

Now we can calculate sin(2a):

sin(2a) = 2 * (5/13) * (12/13)

sin(2a) = (2 * 5 * 12) / (13 * 13)

sin(2a) = 120/169

Next, let's find ctg(2a):

ctg(2a) = cos(2a) / sin(2a)

We can find cos(2a) using the double-angle formula for cosine:

cos(2a) = cos^2(a) - sin^2(a)

cos(2a) = (12/13)^2 - (5/13)^2

cos(2a) = 144/169 - 25/169

cos(2a) = 119/169

Now we can calculate ctg(2a):

ctg(2a) = (119/169) / (120/169)

ctg(2a) = 119/120

Therefore, sin(2a) = 120/169 and ctg(2a) = 119/120.