Here is the detailed deformation process of 4cosx 3sinx 5sin x 5

I'll give you a general formula.

Consider a sinx + b cosx (may wish to set a>0) and mark c = (a 2 + b 2).

then a sinx + b cosx

c (a c sinx + b c cosx) notes (a c) 2 + b c) 2 = 1, so let = arccos(b c).

Then sin = a c

cosθ = b/c

a sinx + b cosx

c(sin sinx + cos cosx) c cos(x- ) can get the form of sin by changing it at the end) here sin and cos in a sinx + b cosx can be turned into cos or sin, the derivation process remains the same, but the way to take it needs to be a little more pondered.

Summary. Hello classmate, can you describe the problem more clearly?

Hello classmate, can you describe the problem more clearly?

Which question?

Question 1. Classmate, your topic is wrong.

What's wrong. If the terminal edge x is on the positive half axis, sin is 0 and cos is 1. The m in the title should also be equal to 0

and the title is not in accord.

This terminal edge should be in the second quadrant in order to satisfy the coordinates of the intersection point of the element circle in the question.

Answer: 3sinx+5cosx=5

Both sides are squared:

9sin²x+30sinxcosx+25cos²x=25………1）

3cosx-5sinx=m

Both sides are squared:

9cos²x-30sinxcosx+25sin²x=m²……2）

1) + (2) get:

m²+25=9+25

m = 9m = -3 or m = 3

So: 3cosx-5sinx=-3 or 3cosx-5sinx=3

Suppose the known condition is 3sinx+5cosx=5

Due to the basic relationship sin 2x+cos 2x=1

And from the world to change with 3sinx+5cosx=5, sinx=5 3(1-cosx).

So there is (5 3) 2*(1-cosx) 2+cos 2x=1, i.e., 17cos 2x-25cosx+8=0

i.e. (17cosx-8) (cosx-1) = 0 so cosx=8 17 or cosx=1

When you guess cosx=8 17, sinx=5 3(1-cosx)=15 17, then 3cosx-5sinx=-3

When cosx=1, it is obvious that Solu sinx=0, and at this time, 3cosx-5sinx=3

Another method: let 3cosx-5sinx=m

Then the above formula is squared on both sides.

9cos 2x-30sinxcosx+25sin 2x=m 2(i) is squared by both sides of the conditional formula.

9sin 2x+30sinxcosx+25cos 2x=25(ii)i)+(ii) and combined with the basic relationship sin 2x+cos 2x=1 have.

m 2 = 9 solution gives m = 3

Let 3cosx-5sinx=m

Then the above formula has two square sides of the ruin.

9cos 2x-30sinxcosx+25sin 2x=m 2(i) by the conditional formula two states meet the square of the side.

9sin 2x+30sinxcosx+25cos 2x=25(ii) book Lu talk i) + (ii) and combine the basic relationship sin 2x+cos 2x=1 have.

m 2 = 9 solution gives m = 3

Solve with a three-function universal formula:

Let tan( 2)=k, from 3sinx+4cosx=5 to get 3*2k (1+k)+4*(1-k) (1+k)=5 to get 9k -6k+1=0

Solution: k=1 3

i.e.: tan = 2k (1-k) = 3 4

3sinx+4cosx=5 squared on both sides.

9sinx^2+24sinx*cosx+16cosx^2=25 (1)

Since sin 2+cosx 2=1; (2) Obtained from (1) (2).

9sinx^2+24sinx*cosx+16cosx^2=25sinx^2+25cosx^2

24sinx*cosx=16sinx 2+9cosx 2 with both sides divided by cosx 2

24tanx=9+16tanx^2

4tanx－3)^2=0

The solution is tanx=3 4

Square on both sides.

9sin 2+24sin*cos+16cos 2=25 (cos 2+sin 2).

24s*c=16c^2+9s^2

Both sides are divided into cosx

24tan=16+9tan^2

(3t-4)^2=0

tanx=4/3

Solution 1: 32sinxcosx is obtained from 3sinx+4cosx=5

sinxcosx

4cosxsinx

cosxsinx

5, i.e. 3?2t

1+t41?t1+t

where t tanx

2, finishing 9t2

6t+1=0, i.e. (3t-1)2

0, and thus t 1

So: tanx 2t1?t

Solution 2: 3sinx+4cosx=5: 5 (3sinx+4

cosx)＝5

Thus sin(x+?1, where tan?＝4

by sin(x+?1 gets: x+?2k + i.e. x 2k + k z so tanx tan (2k + tan (

＝cot?＝3