Given (2sina + COSA) / (sina-3cosa) = - 5, find the value of 3cos2a + 4sin2a

Given (2sina + COSA) / (sina-3cosa) = - 5, find the value of 3cos2a + 4sin2a

Given 2sina + cosa / sina-3cosa = - 5, find the value of 3cos2a + 4sin2a

2sina+cosa=-5sina+15cosa7sina=14cosasina=2cosasina²+cosa²=15cos²a=1cos²a=1/53cos2a+4sin2a=3(2cos²a-1)+8sinacosa=6cos²a-3+16cos²a=22cosa²-3=22/5-3=7/5

To find the value of 3cos2a + 4sin2a by (2sina + COSA) / (sina-3cosa) = - 5

According to the known data, we can sort out and calculate Tana = 2
Tan2a = - 4 / 3 cos2a = 3 / 5 or - 3 / 5 sin2a = - 4 / 5 or 4 / 5
such
3cos2a+4sin2a=±7/5

It is known that sina = 3 / 5 a belongs to (PAI / 2, PAI) Tan (pai-b) = 1 / 2
Finding Tan (a-2b)

Given Sina = 3 / 5 a ∈ (π / 2, π) Tan (π - b) = 1 / 2, find Tan (a-2b) a ∈ (π / 2, π), Sina = 3 / 5 --- cosa = - 4 / 5 --- Tana = - 3 / 4, Tan (π - b) = 1 / 2 --- tanb = - 1 / 2 --- tan2b = (- 1) / (1-1 / 4) = - 4 / 3 --- Tan (a-2b) = (4 / 3-3 / 4) / (1 + 1) = 7 / 24

5x * 2Y = (X-Y) ^ 2 x + 9y = 55 to find X and Y

10xy=55 xy=5.5

To solve a system of linear equations (1) 5x-9y + Z = 2 (2) 5x + 2y-4z = 3 (3) - 5x + Y-Z = 2
Three are connected, and (3) is a negative 5x
Sorry, (1) 5x-3y + Z = 2
That's right. Wrong number

(2) (1) 11y - 5Z = 1; (4)
(3) + (1); - 8y = 4; (5)
Solutions (4) and (5);
Y = - 1 / 2; Z = - 13 / 10;
Substituting (3);
x = 4/25；

If x ＞ y ＞ 0, P = x ^ 3 + 13xy ^ 2 and q = 5x ^ 2Y 9y ^ 3 are the same
q=5x^2*y+9y^3

p-q=x³+13xy²-5x²y-9y³=(x-y)(x²-4xy+9y²)
If x ＞ y ＞ 0, then X-Y ＞ 0, X & # 178; - 4xy + 9y & # 178; = (x-2y) &# 178; + 5Y & # 178; ＞ 0
∴ p＞q

It is known that the inner sides of the triangle ABC are called a, B and C respectively, and a = 2, CoSb = 3 / 5. If B = 4, find the value of sina. (2) if the area of the triangle ABC is 4, find the value of B and C
SINB = 4 / 5, Sina = 2 / 5, don't just copy others?

It is easy to know that the sine of B is 4 / 5, and the sine of a is 2 / 5 obtained from the triangle sine theorem. (2) C is 5 obtained from the formula of area obtained from sine, and then the cosine of B can be obtained from the triangle cosine theorem by solving the equation of B under the root sign

In the triangle ABC, if a = 2, B = 3 and C = 120 ° are known, then the value of sina is in order to find a detailed explanation

The cosine theorem C should give the answer

In the triangle ABC, a is equal to 120 ° B is equal to 1, and the area is root 3, then a + B + C divided by Sina + SINB + sinc equals 3Q

S = bcsina × 1 / 2 = C × radical 3 / 4 = radical 3, so C = 4A = radical (B + c-2bccosa) = 2 radical 3 A / Sina = B / SINB = C / sinc 2 radical 3 / (radical 3 / 2) = 1 / SINB = 4 / sinc, SINB = 1 / 4, sinc = 1, so the original formula = (2 radical 3 + 4 + 1) / (radical 3 / 2 + 1 / 4 + 1) = 4