Sinacosa = 60 of minus 169 Sina + cosa = 7 of minus 13 A is the second quadrant angle. How can we get Sina = 5 of minus 13 cosa = 1 of minus 13

Sinacosa = 60 of minus 169 Sina + cosa = 7 of minus 13 A is the second quadrant angle. How can we get Sina = 5 of minus 13 cosa = 1 of minus 13

In the second quadrant, sin > 0, cos0
sinA=5/13
cosA=-7/13-5/13=-12/13

If we know that sina minus cosa is equal to half, we can find sinacosa

(sina－cosa)²=(1/2)²
sin²a－2sinacosa＋cos²a=1/4
1－2sinacosa=1/4
sinacosa=3/8

The relationship between M and n

sina+cosa=m (sina+cosa) ^2=m^2 1+2sinacosa=m^2 sinacosa=n
1+2n=m^2

Given the vector a = (1,1), B = (2, n), if the absolute value of (a + b) = vector a times vector B, then n =?
(vector a times vector b) that is, a times bcosx

All the following a and B are vectors | a | = √ 2 | B | = √ 4 + n & sup2; a + B = (3, N + 1) a * b = 2 + n left form | a + B | = √ (a + b) & sup2; = √ a | & sup2; + | B | & sup2; + 2 * a * b = √ 2 + 4 + n & sup2; + 2 * (2 + n) = √ n & sup2; + 2n + 10 right form a * b = 2 + n. therefore, the problem is transformed into known √ n & sup2; + 2