Question

Найти sin2a;cos2a;tg2a если sina=-5/13 и cosa<0

Answer 1

Ответ:

Объяснение:

sina=-5/13 => cosa= $-\sqrt{1-sin^{2} a}=- \sqrt{1-25/169} =-12/13$

sin2a=2sina·cosa=2*(-5/13)*(-12/13)=120/169

cos2a= cos²a-sin²a=144/169-25/169=119/169

tg2a=sin2a/cos2a=120/119

Answer 2

To find the values of sin(2a), cos(2a), and tan(2a), we'll use the trigonometric identities:

sin(2a) = 2sin(a)cos(a) cos(2a) = cos^2(a) - sin^2(a) tan(2a) = 2tan(a) / (1 - tan^2(a))

Given that sin(a) = -5/13 and cos(a) < 0, we can find the values of sin(2a), cos(2a), and tan(2a) as follows:

First, let's find cos(a) using the identity cos^2(a) + sin^2(a) = 1: sin^2(a) + cos^2(a) = 1 (-5/13)^2 + cos^2(a) = 1 25/169 + cos^2(a) = 1 cos^2(a) = 1 - 25/169 cos^2(a) = 144/169 cos(a) = -12/13 (since cos(a) < 0)

Now, we can calculate the values of sin(2a), cos(2a), and tan(2a):

sin(2a) = 2sin(a)cos(a) sin(2a) = 2 * (-5/13) * (-12/13) sin(2a) = 120/169

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

tan(2a) = 2tan(a) / (1 - tan^2(a)) tan(2a) = 2 * (-5/13) / (1 - (-5/13)^2) tan(2a) = -10/13 / (1 - 25/169) tan(2a) = -10/13 / (169/169 - 25/169) tan(2a) = -10/13 / 144/169 tan(2a) = -10/13 * 169/144 tan(2a) = -1690/1872

Therefore, the values are: sin(2a) = 120/169 cos(2a) = 119/169 tan(2a) = -1690/1872