Senior 1 Mathematics High School Mathematics Questions Please answer in detail, thank you 5 6 55 34

Solution: This problem can be reduced to sinb-sinc=2sina (root number 3sinc) sinb=sin(180-a-c)=sin(a+c)sin(a+c)-sinc=sinacosc-root number 3sinasinccosasinc-sinc=-root number 3sinasincsinina is not equal to 0

So cosa+ root number 3sina 1

2cos（a－60）＝1

a=120

It's simple, be careful!

This can be reduced to sinb-sinc=2sina (root number 3sinc) sinb=sin(180-a-c)=sin(a+c)sin(a+c)-sinc=sinacosc-root number 3sinasinccosasinc-sinc=-root number 3sinasincsincinc=-root number 3sinasincsinc = 0

so cosa + root number 3sina 1

2cos（a－60）＝1

a=120

sinb-sinc=2sina (root number 3sinc) sinb=sin(180-a-c)=sin(a+c)sin(a+c)-sinc=sinacosc-root number 3sinasinccosasinc-sinc=-root number 3sinasincsincsinc= 0

so cosa + root number 3sina 1

2cos（a－60）＝1

a=120so easy

The first problem is the one above, and then we solve the general formula of an=2 n-1 and then solve the expression of tn, tn=n+2(n-1)+2 2(n-2)+2 n-1 is the sum of the first n terms of a series of differences. The equation is multiplied by 2 at the same time and subtracted by dislocation, I don't know if it's right or not, you can test it.

If you choose 145ab+bc+ac 0, there must be a negative value.

4a b c the smallest must be negative. That is, c0, b<0 can be true.

and a+b+c 0 b+c>-a b+c<0, a<0, then |b+c|<|a|

That is to say|b|<|a|,|c|<|a|

It can be seen that a 2>c 2, in summary: 1, 4, 5 are correct.

From "0" and "0", two of the three real numbers a, b, and c, are negative and one is positive. And》0,4a>b>cSo we can immediately obtain: a>0, b<0, c<0, -a=b+c, so we know 1,4,5 pairs; 2, 3 wrong.

Guarantee the correct !!

1.As can be seen from the question, 1 2(|sin（-x)|)=㏒1/2（|sinx|)

That is, f(-x)=f(x), which can be known as an even function.

2.Let the function f(x) be a periodic function and the minimum period is t, then, f(x)=f(x+t).

i.e. 1 2 (|sinx|)=㏒1/2（|sin（x+t)|t=

The function f(x) is a periodic function, and the period t=

Establish a Cartesian coordinate system, mark the points, you will find ab=5, bc=5, ac=2 3, you can use the cosine theorem to find cosa=(25+12-25) 2*5*2 3= 3 5

sina=√22/5

As you can see from the diagram, if a is obtuse, then c must be to the right of 3.

So ac 2=(c-3) 2+4 2=(c-3) 2+16coaa=(25+c 2-6c+25-c 2) 2ac*ab (-1,0).

Solve (25 3, 17 2).

If nothing else, the master upstairs should have miscalculated, or else he was incompetent.

The cosine theorem gives cosa=(b +c -a ) 2bc)=> cosa= 5 5 (the root of 5 is 5), so sina = (2 5) 5

2) When a is at a right angle, ab is perpendicular to ac according to the slope formula kab*kac=-1, which is calculated as c=25 3

So when c>25 3 the angle a is obtuse.

Hello classmates