Knowing 6 B 4, 3 sin 2 a 2 sin 2 b 2 sina , find the minimum value of sin 2 b 1 2sina

Answer:- 6 b 4

1/2≤sinb≤ √2/2

3sin a-2sin b=2sinasin b=3sin a 2 - sina (3 2) (sina-1 3) -1 6 [-1 2, 2 2]2 3 sina 1 or -1 3 sina 0sin b-1 2sina = 3sin a 2 - sina-1 2sina

3sin²a/2-3sina/2

3/2(sina-1/2)²-3/8

1 3 3 2 (sina-1 2) -3 8 0 or 0 3 2 (sina-1 2) -3 8 2 3

i.e.: -1 3 sin b-1 2sina 0 or 0 sin b-1 2sina 2 3

The minimum value of sin b-1 2sina is -1 3

sin^2b=(3sin^2a-2sina)/2,sin^2 b-1/2sina

3/2(sina-1/2)^2-3/8.

6 b 4, when sina = 1 2, sin 2 b-1 2sina has a minimum value, the minimum value is -3 8

Let f(x)=1

xsinx(0 x 1), then f (x) = xcosx?sinxx2 when 0 x 1, x tanx, f (x)=xcosx?sinxx2tanxcosx?sinx

x20，f（x）=1

xsinx decreases monotonically over the interval (0,1), f(1) f(1 f(15, i.e., sin1

sin1 Shen Zheng sin1

i.e. wide ascension sin1 3sin1

5sin1c b laughing sleepy a so the answer is: c b a

sin(3 +a) = 2sin(3 2+a) i.e. -sina = -2cosa

tana=2

sina-4cosa) (5sina+2cosa) numerator and denominator are divided by cosa.

tana-4) liquid accompaniment (5tana+2).

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TANA=2 YIELDS SINA=2COSAA=(1 2)XSINA

Then combine sin a+cos a=1 ==sin a+1 4(sin a)=1==>sin a=4 5

sin²a+2sinαcosa+2=sin²a+2sinαx(1/2)xsina+2=sin²a+sin²a+2=2sin²a+2

Place sin a = 4 5

Substituting yields sin a+2sin cosa+2=2sin a+2=18 5

sin(3 +a) = 2sin(3 2+a), left = sin(3 2+3 2+a).

sin(3π/2)cos(3π/2+a)+sin(3π/2+a)cos(3π/2)

cos(3π/2+a)

cos(π/2+a)

sina right = 2sin (3 2+a).

2sin(π/2+a)

2cosa, so -sina=-2cosa

tana=2

Is the denominator 5sina+2cosa?

sina-4cosa) (5sina+2cosa) numerator denominator is also punished with cosa.

tana-4)/(5tana+2)

Establish. a=sina,b=sinb,c=sinc

Because the omen is copying.

0-pi 2, the sin function is an increase function, and the maximum value of a corresponds to the maximum value of sina, which is the same reason.

The maximum value of a+b+c also corresponds to the maximum value of a+b+c.

sin²a+sin²b+sin²c=1

So. a²+b²+c²=1

a+b+c)²=a²+b²+c²+2ab+2ac+2bc

Because 2ab<=a +b, 2ac<=a +c, 2bc<=c bridge jujube + b

So. a+b+c)²=a²+b²+c²+2ab+2ac+2bc<=3(a²+b²+c²)=3

a+b+c<=3^

The equal sign is only true when a, b, and c are equal, so when a=b=c=3 there is a maximum, and the angle value for the Tao is at this time.

arcsin

So the maximum value of a+b+c is .

3arcsin

sin²a+sin²b=2sin²c

The concept of the strings is not set by Zhengzai.

a^2+b^2=2c^2

Substituting the cosine theorem:

cosc=(a^2+b^2-

c^2)(2ab)=

c^2(2ab)

So: cosc>0

c is the acute angle of the segment, and the maximum is 90 degrees.

2sin = 3sin stove sells -2sin because - reputation type 6 4 so 0 2sin 1 both 0 3sin -2sin 1

Let x=sin take 0 3x -2x 1 to get -1 3 x 0 or 2 3 x 1

sin -1 2sin =3 2sin -sin -1 2sin =3 2(sin -sin) 3 2(x -x) because -1 3 x 0 or 2 3 x 1

So when x = 2 3, 3 2 (x -x) minimum is -4 27

Because 0<=sinb<=1

So 0<=2sin b<=2;

2sin²b = 2sina-3sin²a;

0<=2sina-3sin a<=2, and -1<=sina<=1;

0<=sina<=2 3; (2sina-3sin a<=2 is constant because 2sina<=2<=2+3sin a).

sin²a+sin²b = sin²a + 1/2*(2sina-3sin²a) = sina-1/2*sin²a=1/2*(1-(sina-1)^2);

Let sina=x; f(x)=1 2(1-(x-1) 2); 0<=x<=2/3;

On the interval [0,2 3], f(x) is a single-drop increment function, so it can be found.

The maximum value is f(2 3) = 4 9; The minimum value f(0)=0;

The value range is [0,4 9].

Let a=sin ,b=sin ,-1<=a<=1,-1<=b<=13a 2-2a+2b 2=0

b^2=(2a-3a^2)/2

Because 0<=b 2<=1

So 0<=(2a-3a 2) 2<=1

The solution is 0<=a<=2 3

So the original formula. a^2+b^2

a^2+(2a-3a^2)/2

a^2/2+a

1/2(a-1)^2+1/2

Therefore, when a=0, the minimum value is -1, 2(0-1), 2+1, 2=0, and when a=2 3, the maximum value is -1, 2(2, 3-1) 2+1, 2=4 9, so the range of values of the original formula is [0, 4, 9].