Cos (a + b) = 12 / 13, cos (2a + b) = 3 / 5, find cosa

Cos (a + b) = 12 / 13, cos (2a + b) = 3 / 5, find cosa

a. B is the acute angle, cos (a + b) = 12 / 13, cos (2a + b) = 4 / 5
Difference angle formula:
cosa=cos[(2a+b)-(a+b)]=cos(2a+b)*cos(a+b)+sin(2a+b)*sin(a+b)
Because a, B
cosa=cos(2a+b)-(a+b)=cos(2a+b)cos(a+b)+sin(2a+b)sin(a+b) =12/13 * 3/5 +_ 5 / 13 * 4 / 5 = 56 / 65 or 16 / 65
Cosa = cos ((2a + b) - (a + b)) can be brought in after it is opened and the corresponding sin (a + b) sin (2a + b) is calculated
Cos (2a + b) = cos (a + A + b) = cosacos (a + b) - sinasin (a + b)
a. B is the acute angle, cos (a + b) = 12 / 13, cos (2a + b) = 4 / 5
Difference angle formula:
cosa=cos[(2a+b)-(a+b)]=cos(2a+b)*cos(a+b)+sin(2a+b)*sin(a+b)
Because a, B
cos(A+(A+B))=4/5
=cosAcos(A+B)-sinAsin(A+B)
=cosA*12/13-sinA*5/13
Next, do it yourself
It is known that ellipse C1 and hyperbola C2 have the same focus F1 and F2, point P is a common point of C1 and C2, △ pf1f2 is an isosceles triangle with Pf1 as the bottom, | Pf1 | = 4, and the eccentricity of C1 is 37, then the eccentricity of C2 is 0______ .
According to the meaning of the question, the eccentricity of C1 is E1 = c1a1 = 2c12a1 = | F1F2 | Pf1 | + | PF2 | = 37, and | Pf1 | = 4, | F1F2 | = | PF2 |, | PF2 | = 3. The eccentricity of hyperbola is E2 = | F1F2 | Pf1 | - | PF2 | = 3
Find the maximum and minimum of the function y = sin square x-cosx
I hope someone can help me by 9:30
Let t = cosx, then - 1 ≤ t ≤ 1, and the function is f (T) = - (T + 1 / 2) ^ 2 + 5 / 4, the opening is downward, the axis of symmetry is - 1 / 2, so the maximum value is f (- 1 / 2) = 5 / 4, and the minimum value is f (1) = - 1
It is known that the ellipse x ^ 2 / M + y ^ 2 / N = 1 and the hyperbola x ^ 2 / p-y ^ 2 / Q = 1 (m, N, P, Q ∈ R +) have the same focus, F1, F2 and P are the intersection points of the ellipse and the hyperbola, then | Pf1 | * | PF2|=
P is on the ellipse
So Pf1 + PF2 = 2 √ M
P on hyperbola
|PF1-PF2|=2√p
PF1-PF2=±2√p
If pf1-pf2 = 2 √ p
PF1+PF2=2√m
PF1=√p+√m
PF2=√m-√p
PF1×PF2=m-p
If pf1-pf2 = - 2 √ p
PF1+PF2=2√m
PF1=√m-√p
PF2=√m+√p
PF1×PF2=m-p
To sum up
PF1×PF2=m-p
The minimum value of function f (x) = sin square x + cosx-1
f(x)=(sinx)^2-cosx-1（x∈R）=1-(cosx)^2-cosx-1=- (cosx+1/2)^2+ 5/4-1,
So when cosx = 1, the minimum value of F (x) is - 2,
So the answer is - 2
Ellipse x ^ 2 / A ^ 2 + y ^ 2 / b ^ 2 = 1 (focus on X axis) and hyperbola x ^ 2 / m ^ 2-y ^ 2 / N ^ 2 = 1 have common focus F1, F2
P is an intersection of them, and we can find the area of △ f1pf2
A.am B.an C.bn D.bm
Simplify radical 2cosx - radical 6sinx=
√2cosx -√6sinx
=√2(cosx -√3sinx)
=2√2[(1/2)cosx -(√3/2)sinx]
=2√2(sinπ/6·cosx -cosπ/6·sinx)
=2√2sin(π/6-x)
This problem can be changed into sine function or cosine function. The expressions are different, but they can be transformed
But the man's answer was wrong
If hyperbola x ^ 2 / 3-y ^ 2 = 1 and ellipse x ^ 2 / 6 + y ^ 2 / 2 = 1 have common focus F1 and F2, and P is an intersection of two curves, then cosf1pf2
According to the definition of ellipse,
There is Pf1 + PF2 = 2 √ 6,
According to the definition of hyperbola,
There are PF2 - Pf1 = 2 √ 3,
So Pf1 = √ 6 - √ 3 PF2 = √ 6 + √ 3
And F1F2 = 4,
So cos ∠ f1pf2 = (Pf1 ^ 2 + PF2 ^ 2-f1f2 ^ 2) / (2 * Pf1 * PF2) = 1 / 3
c=2 m=「6+「3
N = "6 -" 3 "is the root
The cosine theorem shows that (m * 2 + n * 2-2c Square) / 2Mn = 1 / 3 * is the power
Let Pf1 = m, PF2 = n, (M > n) then
m+n=2√6，
m-n=2√3
m=√6+√3，n=√6-√3.2c=4。
Cos ∠ f1pf2 = 1 / 3 obtained from cosine theorem... That's the question I asked two years ago... I used to be a sophomore in high school. Now I am in University. Now I answer... Do you show your brush experience so clearly!
Radical 2cosx - radical 6sinx=
````````
Original formula = 2 √ 2 (cosx * 1 / 2-sinx * √ 3 / 2)
=2√2(cosxcosπ/3-sinxsinπ/3)
=2√2cos(x+π/3)