Solve a trigonometric problem

You make the middle line on the hypotenuse, and you get an angle of 30 degrees.

Let the hypotenuse be a and the height be h, and it is easy to know a=h(t+1 t)....T stands for 15 degrees of tan.

The doubling formula tan30 degrees = 2t (1-t 2) = 1 root 3The solution yields t=2-root3Substituting the * formula yields a=

<> to a function with the same name, and then formulate, you can find the maximum value.

∵f(x)=-2cos²x+2sinx+3=-2(1-sin²x)+2sinx+3

2sin x+2sin +1, no sinx=t [-1,1], then.

y=2t²+2t+1

2(t+1/2)²

The opening of the parabola is upward, and the axis of symmetry t=-1 2 [-1,1],|1/2)-(1)|=1/2﹤|1-(-1 2)丨=3 2

There is a maximum value when t = 1, and 2(1+1 2) = 9 2.

So the maximum value of the function is 9 2.

f(x)=2(sinx+1 2) +1 2,sinx=1,x= 2, the function takes the maximum value.

The maximum value is f(2)=5

Obtained by sin x + cos x = 1:

f(x)=-2(1 - sin²x) +2sinx + 3=-2 + 2sin²x + 2sinx + 3=2sin²x + 2sinx + 1

Considering f(x) as a quadratic function, the formula yields:

2(sin²x + sinx) +1

2(sin²x + sinx + 1/4 - 1/4) +1=2(sinx + 1/2)² 1/2 + 1=2(sinx + 1/2)² 1/2

The range of the sinusoidal function is [-1,1].

1/2≤sinx + 1/2≤3/2

When sinx=1, f(x) is the maximum.

i.e.: f(x)=2 (3 2) 1 2=5

∵sin(π-a) -cos(π+a)

sina + cosa

Root number 2) sin(a + 4) = (root number 2) 3 sin(a + 4) = 1 3

sina - cosa = root number 2) sin( 4 - a) = root number 2) cos[ 2 - ( 4 - a)] root number 2) cos( 4 + a) = 4 3

(1) As long as sinx-cosx>0, the auxiliary angle formula solves sinx-cosx=root 2sin(x-4)>0 Note that this is a minus, not a plus.

Do the rest yourself.

2) As long as the true number is a periodic function, the minimum positive period is 2

Hello, the problem can be 1) as long as sinx-cosx>0, solve sinx-cosx=root 2sin(x+4)>0 by the auxiliary angle formula and do it yourself.

2) As long as the true number is a periodic function.

Therefore, the minimum positive period is 2

(1) y satisfies the odd-even function condition f(x)=f(-x) suoyi sin(wx+ )sin(-wx+ )

f(x)=-f(-x) sin(wx+α)sin(-wx+α)

suiyi sin(wx+α)sin(-wx+α)sin(-wx+α)2sin(-wx+α)0

sin(wx+ )sin(-wx+ )0 suoyi wx+a=k -wx+a=k add a=k

2) In the same way, y=cos(wx+ )y=cos(-wx+ )y=-cos(-wx+ ) to get cos(-wx+ )0=cos(wx+ )

wx+ =2k -wx+ =2k is added to give a=2k

3 ） tan(wx+α）tan（-wx+α）0 wx+α=kπ -wx+a=kπ a=kπ

I don't know if that's the case, it's not easy to write so much.

Direct plotting The sine and cosine function has odd (even) properties with only 1 4 periods of translation.

a=t/4=2kπ /w / 4= kπ /2w

(x^2-1)sinb-(x^2-x)sinc-(x-1)sina=0

i.e. (sinb-sinc) x 2 + (sinc-sina) x + (sina-sinb) = 0;

(sinc-sina)^2-4(sinb-sinc)(sina-sinb)

sina)^2+4(sinb)^2+(sinc)^2-4sinasinb+2sinasinc-4sinbsinc

sina-2sinb+sinc)^2

Since the equation (x 2-1)sinb-(x 2-x)sinc-(x-1)sina=0 has two equal real roots, so δ=0, then sina-2sinb+sinc=0, according to the sine theorem there is a-2b+c=0, so a, b, c are equal difference series.

tan(a 2) cannot be determined, and its value range is: (0, 3 3).

f(x)=sin2wx+ 3sinwxsin(wx+ plexus modulo union2)1 2(infiltrate 1-cos2wx)+ 3sinwxcoswx1 2-1 2 cos2wx + 3 2 sin2wx1 2-(sin 6cos2wx-cos 6sin2wx)1 2-sin( 6-2wx).

sin(2wx-π/6)+1/2

So the minimum positive period of the coder is 2 2w= w= , so the value of w is 1