If the inner angles a, B and C in the triangle ABC form an arithmetic sequence and the tolerance D ＞ 0, the opposite sides of a, B and C are a, B and C respectively 1、 If C = 2A, find the value of D. 2. For the function f (d) = sin "a + cos" C, find the range of F (d)

If the inner angles a, B and C in the triangle ABC form an arithmetic sequence and the tolerance D ＞ 0, the opposite sides of a, B and C are a, B and C respectively 1、 If C = 2A, find the value of D. 2. For the function f (d) = sin "a + cos" C, find the range of F (d)

1、 D = 30 ° 2: the range of F (d) is (- 0.5, (1 + radical 3) / 2) by limit method

Triangle ABC, known as cosacosb, sinasinb, judge the shape of triangle ABC

Right angle

In triangle ABC, sinasinb ＜ cosacosb is known. Try to judge the shape of triangle ABC

Cos (a + b) > 0
A + B 90 degrees
Obtuse triangle

In triangle ABC, 2BC * cosa = radical 3 (c * cosa + A * COSC), find the value of A. if a = 2C = 2 radical 3 and B is greater than C, find the area of triangle

I think there is something wrong with this topic. The topic should be 2B * cosa = root 3 (c * cosa + A * COSC), otherwise a cannot be determined. I will solve this problem according to my understanding. (1) according to the sine theorem, there is a = 2rsina B = 2rsinb C = 2rsinc2b * cosa = √ 3 (c * cosa + A * COSC) 2 * 2rsinb * cosa = √ 3 (2rsi

Let 2 sin square a be cosa = 2 in the acute triangle ABC, and find the size of angle A

2(sinA)^2+cosA=2
2(1-(cosA)^2)+cosA=2
2-2(cosA)^2+cosA=2
cosA(2cosA-1)=0
cosA=0
cosA=1/2
Because it's an acute triangle
So: cosa = 1 / 2
A=60

In △ ABC, we know BC = 5, sinc = 2sina, then ab=______ ．

Using the sine theorem to simplify sinc = 2sina, we get: ab = 2BC, ∵ BC = 5, ∵ AB = 25

In △ ABC, BC = 5, AC = 3, sinc = 2sina

(1) In △ ABC, from the sine theorem, we can get bcsina, abbc = & nbsp; sincsina = 2, ｜ AB = 2 × BC = 25. (2) in △ ABC, from the cosine theorem, we can get BC2 = AB2 + ac2-2ab · AC · cosa, 5 = 20 + 9-125cosa, ｜ cosa = 255, ｜ Sina = 1 − cos2a = 55

In △ ABC, if ACOS & # 178; C / 2 + cos & # 178; a / 2 = (3 / 2) B is satisfied, then a + C = 2B, B = π / 4, B = 2, and s △ ABC is obtained

(1) You've missed the title. There's a C in front of a / 2
acos²(C/2)+ccos²(A/2)=(3/2)b
a(1+cosC)/2+c(1+cosA)/2=(3/2)b
a(1+cosC)+c(1+cosA)=3b
From the sine theorem
sinA(1+cosC)+sinC(1+cosA)=3sinB
sinA+sinAcosC+sinC+sinCcosA=3sinB
sinA+sinC+sin(A+C)=3sinB
sinA+sinC+sinB=3sinB
sinA+sinC=2sinB
From the sine theorem
a+c=2b
(2)
From cosine theorem
cosB=(a²+c²-b²)/(2ac)
[(a+c)²-2ac-b²]/(2ac)=cosB
[(2b)²-2ac-b²]/(2ac)=cos(π/4)
(3b²-2ac)/(2ac)=√2/2
B = 2, put in, sort out, get
(2+√2)ac=12
ac=12/(2+√2)=12(2-√2)/2=6(2-√2)
S△ABC=(1/2)acsinB
=(1/2)·6(2-√2)·sin(π/4)
=(1/2)·6(2-√2)·(√2/2)
=3√2-3

cos (180/5)* cos(360/5)=?

cos (180/5)* cos(360/5)
=cos 36* cos72
=cos 36*sin 36*cos72/sin 36
=1/2sin72*cos72/sin 36
=1/4sin144/sin 36
=1/4sin36/sin 36
=1/4

In △ ABC, the opposite sides of angles a, B and C are a, B, C, a = 2 √ 3, B = 2, ∠ B = 30 °, C = 2 √ 2, cosa = - 1 / 2
(1) If f (x) = cos2x + csin ^ 2 (x + b), find the minimum positive period and simple increasing interval of F (x)

According to the meaning of the title, the size of a is 120 degrees, so the angle c is 30 degrees, so C = 2?