Trigonometric Synthesis Questions, Trigonometric Synthesis Questions

y=sinx and y=sin|x|The image is symmetrical with respect to the y-axis.

When x>=0, sin|x|=sinx

When x>=0, sin|x|=sin(-x)=-sin(x) Therefore, the original proposition is a false proposition.

y=sinx and y=|sinx|The image is symmetrical with respect to the x-axis.

sinx|The lower half of the sinx image (quadrants iii and iv) is symmetrical to the top half of the x-axis. And the upper part is the same for both.

So, the original proposition is false.

3) Correct;

4) Correct.

I wish you progress in your studies and go to the next level!

If you don't understand, please ask in time, satisfied, o( o thank you.)

Solution: The false proposition is

The image of y=sinx and y=sin x is symmetrical with respect to the y-axis: a true proposition. When x 0 y=sin x = sinx, i.e. at x 0.

The image of y=sinx coincides with y=x; And at x 0 y=sin x x =sin(-x)=-sinx, so at x 0 y=sinx and y= x the image is symmetrical with respect to the x-axis. Taken together, the images of y=sinx and y=sin x are symmetrical with respect to the y-axis.

The image of y=sinx vs. y=sinx is symmetrical with respect to the x-axis: a false proposition. Because the image of y= sinx is the image of y=sinx.

The image below the x-axis is flipped 180 degrees with the x-axis as the axis, and all are shifted to the top of the x-axis; Therefore, it is a false proposition.

The image of y=cosx is different from y=cosx: false proposition. Because when x 0 y=cos x =cosx; and x 0 y=cos x =cos(-x)=cosx; Therefore, the image of y=cosx and y=cosx is the same.

The image of y=cosx is the same as y=cosx: false proposition.

The image of y=sinx and y=-sin(-x) is symmetrical with respect to the origin: true proposition. Since y=-sin(-x)=sinx, the image is y=sinx.

image, while the image of y=sinx is symmetrical with respect to the origin.

You've made the right choice!

1) Simultaneous y=cos, y=1+1 2sin2, eliminate y has:

cos²α=1+1/2sin2α

1+cos2α）/2=1+1/2sin2α1+cos2α=2+sin2α

cos2α-sin2α=1

cos2αcosπ/4-sin2αsinπ/4=√2/2cos（2α+π/4）=√2/2

[0, 4], 2 + 4 [ 4,3 4]f(x)=cos x=(1+cos2x) 2y=cosx, the axis of symmetry is: y=k, k z, f(x) is: 2x=k, x=k 2, k z

g（2x）=g（kπ）=1+1/2sin2kπ=1h(x)=f(x)+g(x)=cos²x+1/2sin2x=（1+cos2x）/2 +1/2sin2x

cos2x+sin2x）/2 +1/2

sin 4cos2x+cos 4sin2x) 2 2+1 2( 2 2)sin(2x+ 4)+1 2 x [0, 4], 2x+ 4 [0,3 4],sin(2x+ 4) [2 2,1].

The range of 2 2 * 2 2 + 1 2 h(x) 2 2*1 +1 2h(x) is: [1, ( 2+1) 2].

Solution: (1) f( )=g( )

1 2cos2 -1 4sin2 = 1 2 (root 5) 4 [2 (root 5) cos2 -1 (root 5) sin2 ] = 1 2

cos(2 + =2 (root 5) (let cos = 2 (root 5))2 = 2k (k n).

kπ （k∈n）

2）x0=kπ（k∈n）

g(2x0)=1+1 2sin4x0=1+1 2sin(4k)=13)h(x)=cos2x+1 2sin2x+1 (root5) 2[2 (root5)cos2x+1 (root5)sin2x]+1(root5) 2sin(2x+ )1 (let sin =2 (root5) and (0, 2)).

2x∈[0，π/2]

h（x）∈[3／2，2]

[[[1]]]

1]∵cosx=1-2sin²(x/2)∴2sin²(x/2)=1-cosx

4sin²(x/2)=2-2cosx

2]cos2x=2cos²x-1

3] Function f(x) = 2-2cosx + 2cos x-1 + (1 2) = 2cos x-2cosx + (3 2).

It can be obtained from the inscription.

f(a)=2cos a-2cosa+(3 2)=1 Multiply both sides of the above equation by 2 to obtain.

2cosa-1)²=0

cosa=1/2

It can be obtained from the cosine theorem.

1/2=cosa=(b²+c²-a²)/(2bc)∴bc=b²+c²-a²

b²+c²-3

b²+c²-bc=3

It is derived from the basic inequality.

b²+c²≥2bc

b²+c²-bc≥bc

It can be obtained by combining B+C-BC=3.

bc 3, easy to know, equal sign is obtained only when b = c = 3.

bc)max=3

At this point, b=c=3

In this case, abc is an equilateral triangle with side length = 3

At this time, the area s=(3, 4), (3).

①sin^2(a/2)=(1-cosa)/2，cos(2a)=2cos^2(a)-1

f(a)=4sin^2(a/2)+cos(2a)+1/2=2(1-cosa)+2cos^2(a)-1+1/2=1

Let cosa=x, there are 2x 2-2x+1 2=0, i.e., (x-1 2) =0, x=cosa=1 2, a=60°(0 a 180°).

From the cosine theorem, 2bc*cosa=b +c -a, i.e., bc=b +c -3

The transformation yields b + c -2bc=(b-c) =3-bc, bc=3-(b-c), bc takes the maximum value, then b=c, bc=3

s△=1/2*bc*sina=1/2*3*√3/2=3√3/4

f(x)=4sin^2（x/2）+cos2x+1/2=4*(1-cosx)/2+2(cosx)^2-1+1/2=2(cosx)^2-2cosx+3/2

f(a)=2(cosa)^2-2cosa+3/2=14(cosa)^2-4cosa+1=0

2cosa-1)^2=0

cosa=1/2

Because 0a = 60°

By the sinusoidal theorem, b sinb=c sinc=a sinina=3 b sinb*c sinc2*2

bc=4sinbsinc=-4[cos(b+c)-cos(b-c)]/2=-2[cos(180°-a)-cos(b-c)]=2cos(b-c)+1

Cause|b-c|<180°-a=120°

When b-c=0, the maximum value is bc=3

s abc=1 2bcsina=3 4*3 three-quarters the root number three).

cos(π/4-a)=2√2·cos2a

cos(π/4)cosa+sin(π/4)sina=2√2(cos²a-sin²a)

2/2)(cosa+sina)

2 2· (cosa+sina) (cosa-sina) about to go to cosa+sina, easy to get.

cosa-sina=1/4

Squared, got. 1-2sinacosa=1/16

So sin2a = 15 16

f(x)=√3sin2x+cos2x

f(z)=√3sin2z+cos2z=5/6√3sin2z=5/6-cos2z

Square 3sin 2z = 3-3 cos 2z = 25 36-5 3 * cos2z + cos 2z

cos 2z-5 12*cos2z-83 144=0z belongs to [v4, v2], 2z belongs to [vultures 2, v], so cos2z<0

So cos2z=(5 12- 357 12) 2 i.e. cos2z=(5- 357) 24

The equation of doubling angle is used to convert f(x) to f(x)=2cos(2x-3), which is set by the problem, 2cos(2z- 3)=6 5, because z belongs to [V4, Vulture 2], so sin(2z- 3)=4 5; and because cos(2z) = cos((2z- 3) + 3); From the formula, cos(2z) = (3-4 * root number 3) 10;