sin3x/sinx - cos3x/cosx = 2?

Question

how do you figure that one out?

its high school pre-calc trigonometry

Answer 1

sin3x/sinx - cos3x/cosx = 2
(3sin x -4sin^3 x)/sin x-(4cos^3 x-3 cos x)/cos x=2
3-4sin^2 x-(4cos^2 x-3)
6-4(sin^2 x+cos^2 x)=2
6-4=2
2=2

Answer 2

sin2x = sin(3x-x) = sin3xcosx - sinxcos3x (1) sin2x = 2sinxcosx (2) (1) = (2) then 2sinxcosx = sin3xcosx - sinxcos3x if sinxcosx <>0 then 2 = [sin3xcosx - sinxcos3x] / sinxcosx 2 = sin3x/sinx - cos3x/cosx Voila.

Answer 3

Most of the preceding answers (3/5) seem to show
the harder "triggering" work of applying the appropriate
trig formula, BUT I think it's critical to point out that the left side of the equation always yields 2 ONLY as long as it is defined. You need to throw out all x that provoke denominators of zero:
sin x = 0 when x = n(pi),
cox x = 0 when x = (2n + 1) pi/2
n is any integer.
A little further study/organizing and you could combine the 2 conditions: x CAN"T equal n (pi/2)

Answer 4

You have to know multiple angle formulae.
Sin3x = 3Sinx - 4(Sinx)^3
Cos3x = 4(Cosx)^3 - 3Cosx

Given:
Sin3x/Sinx - Cos3x/Cosx
= [3Sinx - 4(Sinx)^3]/Sinx - [4(Cosx)^3 - 3Cosx]/Cosx
= [3 - 4(Sinx)^2] - [4(Cosx)^2 - 3]
= 3 - 4(Sinx)^2 - 4(Cosx)^2 + 3
= 6 - 4(Sinx)^2 - 4(Cosx)^2

Answer 5

sin3x/sinx - cos3x/cosx = 2
<=> sin3x*cosx - cos3x*sinx = 2sinx*cosx
<=> sin(3x - x) = sin2x
<=> sin2x = sin2x

Answer 6

Operate on the left hand side first
Use sin(3x) = sin (2x +x) = sin 2x cos x +cos 2x sin x
Now substitute 2sinx cos x for sin 2x and 1-sin^2 x for cos x;
sin3x= 2sinx cos^2 x + sin x (1-sin^2x)
Divide by sin x getting (sin 3x)/sin x= 2cos^2 x +1 - sin^2x

Now operate similarly on the right hand side;
Use cos 3x = cos(2x +x)=cos 2x cos x - sin 2x sin x
cos 3x= (1-sin^2x)cosx - 2sin^2 x cosx
(cos 3x)/cos x = 1-sin^2 x - 2sin^2 x

(sin 3x)/sin x - (cos 3x)/cos x =
2cos^2 x +1 -sin^2 x -1 +sin^2x +2sin^2x
=2cos^2x +2 sin^2 x =2(cos^2x +sin^2 x) = 2

Answer 7

do it yourself. look in the book. It's the only way.Life is never easy. But looking back at high school I know I did the right now because I majored in statistics at Harvard

Answer 8

sin3x/sinx-cos3x/cosx
(3sinx-4sin^3x)/sinx-(4cos^3x-3cosx)/cosx
3-4sin^2x-4cos^2x+3
=6-4(sin^2x+cos^2x)
=6-4
=2

Answer 9

(sin(3x)/sin(x)) - (cos(3x)/cos(x)) = 2

sin(3x) =
sin(2x + x) =
sin(2x)cos(x) + cos(2x)sin(x) =
2sin(x)cos(x)cos(x) + (cos(x)^2 - sin(x)^2)sin(x) =
2sin(x)cos(x)^2 + sin(x)cos(x)^2 - sin(x)^3 =
3sin(x)cos(x)^2 - sin(x)^3

cos(3x) =
cos(2x + x) =
cos(2x)cos(x) - sin(2x)sin(x) =
(cos(x)^2 - sin(x)^2)cos(x) - 2sin(x)cos(x)sin(x) =
cos(x)^3 - cos(x)sin(x)^2 - 2cos(x)sin(x)^2 =
cos(x)^3 - 3cos(x)sin(x)^2

So now we have

(sin(3x)/sin(x)) - (cos(3x)/cos(x)) = 2

((3sin(x)cos(x)^2 - sin(x)^3)/sin(x)) - ((cos(x)^3 - 3cos(x)sin(x)^2)/(cos(x)) = 2

(3cos(x)^2 - sin(x)^2) - (cos(x)^2 - 3sin(x)^2) = 2

3cos(x)^2 - sin(x)^2 - cos(x)^2 + 3sin(x)^2 = 2

2cos(x)^2 + 2sin(x)^2 = 2

2(cos(x)^2 + sin(x)^2) = 2

2(1) = 2

2 = 2

Answer 10

Is this high school math, geometry?

Good luck. Thanks.