It is known that 5 13sinX 12 13cosX 0 6 is found cosX

Original -5 13sinx+12 13cosx=-5sinx+12cosx=

Let x=90'+x

So -5sin(90+x)+12cos(90+x)=finish-5cosx-12sinx=

1)-60sinx+144cosx=

2)-60sinx-25cosx=39

Synopid: cosx=

Original -5 13sinx+12 13cosx=This formula is sub-formed: asinx+bcosx= (A 2+b 2)cos(x+ )

So the original formula is equal to (-5 13) 2+(12 13) 2cos(x+ )=cos(x+ )

tanφ=b/a

And then it should be able to do it too.

Because (-5 13) 2+(12 13) 2=1, sin = 12 13, cos = -5 13, ( 2

So -5 13sinx+12 13cosxcos sinx+sin cosx

sin（α+x）=

Thus, cos( +x)=+- 1-sin( +x) 2]cos =cos[( x)-

cos(+x)cos+sin(+x)sin -56 65 or -16 65

1/(sin²xcos²x)dx

4/(4sin²xcos²x)dx

1/sin²2xdx

csc²2x

d(2x)-2cot2x

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Because (sinx) 2+(cosx) 2=1 is good to consume (-cosx-7 13) 2+(cosx) 2=12(cosx) 2+14 13cosx-120 bridge 169=0, so cosx=-12 Youshan 13

sinx=5/13

cosx+2sinx=-12/13+10/13=-2/13

Original formula = 1 (1+(cosx) 2) dx numerator and denominator divided by (cosx) 2

(secx) 2 ((secx) 2+1) dx= 1 ((secx) 2+1) d (tanx) = 1 ((tanx) 2+2) d (tanx) set of formulas = 1 2*arctan((tanx) 2)+c