High 1 trigonometric function problem cosx cosy cosxcosy

By the product sum difference formula.

cosx+cosy=2*cos((x+y)/2)*cos((x-y)/2)

by and difference product formulas.

cosxcosy=1 2*[cos(x+y)+cos(x-y)]Therefore, 2*cos((x+y) 2)*cos((x-y) 2)=1 2*[cos(x+y)+cos(x-y)]

i.e. 2*so cos((x-y) 2) = another round off)

Solution: cosx+cosy=2cos((x+y) 2)*cos((x-y) 2).

cosxcosy=1/2[cos(x+y)+cos(x-y)]cos²((x+y)/2)+cos²((x-y)/2)-1 ②cos((x+y)/2)= ③

Simplified:

cos((x-y) 2) cos((x-y) 2), the solution gives cos((x-y) 2)=, or cos((x-y) 2)=

1≤ cos((x-y)/2) ≤1

cos((x-y) 2) = rounded, so cos((x-y) 2) = description: product sum difference formula:

sin sin =-[cos( +cos( - 2cos cos =[cos( +cos( - 2sin cos =[sin( +sin( - 2cos sin = [sin( +sin( - 2 and the difference product formula:

sin +sin =2sin[(2]cos[(2]sin -sin =2cos[(2]sin[(2]cos +cos =2cos[(2]cos[(2]cos -cos =-2sin[(2]sin[(2]sin[(2]sin[(2]double angle formula: cos =2cos (2)-1

Upstairs I got the answer wrong and got the "product sum difference" formula wrong, and my answer is as follows:

Solution: Because cosx+cosy=cosxcosy, 2cos[(x+y) 2]cos[(x-y) 2]=(1 2)[cos(x+y)+cos(x-y)]-i), which is known by the double angle formula, cos(x+y)=2 2-1, cos(x-y)=2cos 2-1---ii), is brought into the above equation and obtained.

2cos[(x+y) 2]cos[(x-y) 2]=(1 2)[2 2-1+2cos 2-1], known cos((x+y) 2)=, then.

2*, i.e., cos, cos[(x-y) 2], and the solution gives cos[(x-y) 2]= (cos[(x-y) 2]=rounded)--iii).

That's it!

Note: (i) This step is utilization"and the difference product "cosx+cosy=2cos[(x+y) 2]cos[(x-y) 2]), which is replaced with the formula "product sum difference" cosxcosy=(1 2)[cos(x+y)+cos(x-y)];

ii) is the use of the double angle formula cos2x=2(cosa) 2-1, where () 2 refers to the square within ();

iii) because cos[(x-y) 2] is a trigonometric function, and its value range is [-1,1], which is not in this range, so it is discarded;

In addition, the penultimate and third-to-last steps in the answer can be omitted in the answer. 】

It's the first time I've answered the question, maybe it's a mistake to write any symbols, so take a closer look yourself!

cosx+cos2x+cos3x+.+cosnx

cosx+cos2x+cos3x+.+cosnx=sin(x/2)*[cosx+cos2x+cos3x+.+cosnx] sin(x 2) (Move sin(x 2) into square brackets and simplify).

[2sin(x/2)]

Isoangular trigonometric functions.

1) Squared Relationship:

sin^2(α)cos^2(α)=1

tan^2(α)1=sec^2(α)

cot^2(α)1=csc^2(α)

2) Product Relationship:

sinα=tanα*cosα cosα=cotα*sinαtanα=sinα*secα cotα=cosα*cscαsecα=tanα*cscα cscα=secα*cotα

cosx+cos2x+..cosnx

1/2[(cosx+cosnx)+(cos2x+cos(n-1)x)+.cosnx+cosx)] Multiply by 2 and then divide by 2

cos(n+1)x/2][cos((n-1)x/2)+cos(((n-3)x/2)+.cos((n-(2n-1))x2) and the differential product.

cos(n+1)x/2/sin(x/2)]*sin(x/2)*cos((n-1)x/2)+sin(x/2)*cos(((n-3)x/2)+.sin(x 2)*cos((n-(2n-1))x 2) multiply sin(x 2) and divide by sin(x 2).

1/2[cos(n+1)x/2/sin(x/2)][sin(nx/2)+sin((2-n)x/2)+sin((n-2)x/2)+sin((4-n)x/2)+.sin((2-n)x 2)+sin(nx 2)] product sum difference.

[sin(x 2)]

There is one more way:

The sum and difference product formula can be further simplified to obtain the final result: [sin(x 2)].

These two methods use the product sum difference sum and difference product formulas, as long as you flexibly master these two types of formulas, it is easy to do.

Here are some of the main formulas used:

Let t=sinx+cosx= 2sin(x+ 4)t [-2, 2] and t ≠-1

sinx+cosx)²=1+2sinxcosxsinxcosx=(t²-1)/2

y=sinxcosx (1+sinx+cosx)[(t Li-1) 2] (1+t)=(t-1)(t+1) [2(1+t)].

t-1)/2

t [-2, 2] 2-1 t-1 2-1-( 2+1) (t-1) 2 ( 2-1) 2t≠-1 y≠zhencha-1

The range of y is [-(2+1) 2,-1)u(-1,( 2-1) 2],8,

Solution: cosx+cosy=2cos((x+y) 2)*cos((x-y) 2).

cosxcosy=1/2[cos(x+y)+cos(x-y)]

cos²((x+y)/2)+cos²((

x-y)/2)-1

cos((x+y)/2)=

Simplified:

cos²((

x-y)/2)

cos((x-y)/2)

The solution yields cos((x-y) 2)=, or cos((x-y) 2)=

cos((x-y)/2)≤1∴

cos((x-y) 2) = rounded, so cos((x-y) 2) =

Description: Accumulation and Difference Formula:

sinαsinβ=-[cos(α+cos(α-/2

cosαcosβ=[cos(α+cos(α-/2

sinαcosβ=[sin(α+sin(α-/2

cosαsinβ=[sin(α+sin(α-/2

and the difference product formula:

sinθ+sinφ=2sin[(θ/2]cos[(θ/2]

sinθ-sinφ=2cos[(θ/2]sin[(θ/2]

cosθ+cosφ=2cos[(θ/2]cos[(θ/2]

cosθ-cosφ=-2sin[(θ/2]sin[(θ/2]

Doubling formula: cos = 2cos (2)-1

cosx-cosy=1 2 sinx-siny=-1 3 are squared on both sides.

cosx) 2-2cosxcosy+(cosy) 2=1 4(sinx) 2-2sinxsiny+(siny) 2=1 9 add and (cosx) 2+(sinx) 2=1 (cosy) 2+(siny) 2=1

2-2(cosxcosy+sinxsiny)=1/4+1/9=13/36

cos(x-y)=cosxcosy+sinxsiny=59/72,9,