High School Math Trigonometry

1.Left = tan -sin = sin (sec -1) = tan * sin = right.

2.Left = cos -2cos +1 + sin = 2-2cos = right.

3.Left-right=sin quad x-2sin xcos x +cos quad x-1=(sin x+cos x) 1=0

Get sin = 2cos to bring in.

Derivation = 3

1 sin²α/cos²α-sin²α=(sin²α-sin²αcos²α)/cos²α=sin²α(1-cos²α)/cos²α

sin * sin cos = result.

Remove the parentheses cos +1-2cos +sin = 2-2cos 3 sin to the fourth power x + cos to the fourth power x = (sin x + cos x) -2sin xcos x

1-2sin²xcos²x

tan =2 yields sin =2 root number 5 cos =1 root number 5sin +cos sin -cos =2 root number 5 + 1 2-1 root number 5

1 root number 5+1 2

2..（cosβ-1）²+sin²β=cos²β-2cosβ+1++sin²β=2-2cosβ(cos²β+sin²β=1)

sin²x+cos²x)²-2sin²xcos²x1-2sin²xcos²x

The numerator and denominator are divided by cos).

tanα+1/tanα-1=2+1/2-1=3

f(x)=(1+cotx)(sinx^2)-2sin(x+π/4)sin(x-π/4) =1+cosx/sinx)(sinx^2)+2sin(x+π/4)sin(π/4-x)

1+cosx/sinx)(sinx^2)+2cos(π/4-x)sin(π/4-x)

sinx^2+sinxcosx+sin(π/2-2x)=sinx^2+1/2sin2x+cos2x=(1-cos2x)/2+1/2sin2x+cos2x

1 2 (sin2x+cos2x)+1 2=(root number 2x) sin(2x+ 4) 2+1 2

Because x [ 12, 2] , we get 2x+ 4 [5 12,5 4], sinx [-2,1], so that the original value range [0,1].

If there is no error in the problem, the solution: because a=b, then a=b, a-c= 3, and because 3c+2 3=, so c= 9,a=b=4 9,You can get it with a calculator or a lookup table.

The revised answer should look like this.

a+c=2b is known by the sinusoidal theorem that sina + sinc = 2sinb, that is, 2sin[(a+c) 2]cos[(a-c) 2]=4sin(b 2)cos(b 2), and because cos(b 2),=2sin[(a+c) 2], a-c = 3, 2sin(b 2) = root number 3 divided by 4,0< b 2<2 parts, cos(b 2) = root number 13 divided by 4, so sinb = root number 39 divided by 8< p>

a=b, then a=b, a-c= 3,2a+c=

Get c= 9, a=b=4 9 so sinb=sin4 9, use the calculator.

Let k=sinx-cosx

Then 2sinxcosx=k 2-(sinx) 2-(cosx) 2 sinxcosx=(k 2-1) 2

and k=sinx-cosx= 2sin(x-45°) 1, 2](x [0, ]).

f(k)=(k^2-1+2k)/2

(k+1) 2-2] 2 [-2,(1+2 2) 2] its maximum value is (1+2 2) 2

The minimum value is -2

sina > 0 and cosa < 0

So a is in the second quadrant.

Because sin3 4 = 2 2

So a=3 4+2k, where k z

Answer: When you go to college, you will know a lot about the application of trigonometric functions, complex variable functions, three-phase alternating current, vector spaces, the relationship between spatial planes and straight lines, and the substitution of variables in calculus, etc., all of which are trigonometric functions. The construction of bridges in real life, the cantilever beams of cranes, etc., are inseparable from the application of trigonometric functions. Therefore, trigonometric functions are the basics and can lay a good foundation for further study.

y＝sin（2x-π/3)=cos(2x-π/3-π/2)=cos(2x+π/2-4π/3)=cos[2(x+π/4)-4π/3]

So just move the image of the function y=cos(2x-4 3) to the left by 4

Look at the picture, so the answer is to shift the vultures 4 to the left, you should be a sophomore in high school by now, so you can take a look.