Algebra help - on calculus problem?

Question

I'm tutoring a calculus student, and the current problem I'm helping her with is the area between the curves of 
y = sin3x   and  y = cos2x  (from x = 0 to x = 1)

The integral isn't really a problem - - except I need to determine where
the intersection is  (ie:  sin3x = cos2x  where 0 < x < 1)

I tried using the double angle identities with the sin (A + B) identity, but either I went the wrong direction, or this is the wrong method to use.  
Any help??  (Just need the proper method - a little guidance wouldn't hurt, though)

Brian · Accepted Answer

Note that sin(A) = cos((pi/2) - A), so we have

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

(pi/2) - 3x = 2x  ---->  pi/2 = 5x  ---->  x = pi/10.

From x = 0 to x = pi/10 we have cos(2x) >= sin(3x) and

from x = pi/10 to x = 1 we have sin(3x) >= cos(2x).

So the area will be

integral(x=0 to pi/10)((cos(2x) - sin(3x)) dx) +

integral(x = pi/10 to 1)((sin(3x) - cos(2x)) dx) =

((1/2)*sin(2x) + (1/3)*cos(3x)) (from x = 0 to x = pi/10) +

((-1/3)*cos(3x) - (1/2)*sin(2x)) (from x = pi/10 to x = 1) =

[(1/2)*sin(pi/5) + (1/3)*cos(3pi/10) - (0 + (1/3))] +

[(-1/3)*cos(3) - (1/2)*sin(2) - ((-1/3)*cos(3pi/10) - (1/2)*sin(pi/5))] =

sin(pi/5) + (2/3)*cos(3pi/10) - (1/3)*cos(3) - (1/2)*sin(2) - (1/3) =

(5/3)*sin(pi/5) - (1/3)*cos(3) - (1/2)*sin(2) - (1/3) = 0.52166 to 5 decimal places,

since cos(3pi/10) = sin(2pi/10) = sin(pi/5).

JoAn · Answer

y = (sin 3x) and y = (cos 2x)


(sin 3x) = (cos 2x)


note: (sin 3x) = (sin (x + 2x))
note: (cos 2x) = 1 - 2(sinÂ²x)


(sin (x + 2x)) = 1 - 2(sinÂ²x)


note: (sin (x + 2x)) = (sin x)(cos 2x) + (cos x)(sin 2x)


(sin x)(cos 2x) + (cos x)(sin 2x) = 1 - 2(sinÂ²x)

(sin x)(1 - 2(sinÂ²x)) + (cos x)(sin 2x) = 1 - 2(sinÂ²x)

((sin x) - 2(sinÂ³x)) + (cos x)(sin 2x) = 1 - 2(sinÂ²x)


note: (sin 2x) = 2(sin x)(cos x)


((sin x) - 2(sinÂ³x)) + (cos x)(2(sin x)(cos x)) = 1 - 2(sinÂ²x)

((sin x) - 2(sinÂ³x)) + 2(sin x)(cosÂ²x) = 1 - 2(sinÂ²x)


note: (cosÂ²x) = 1 - (sinÂ²x)


((sin x) - 2(sinÂ³x)) + 2(sin x)(1 - (sinÂ²x)) = 1 - 2(sinÂ²x)

((sin x) - 2(sinÂ³x)) + (2(sin x) - 2(sinÂ³x)) = 1 - 2(sinÂ²x)

(sin x) - 2(sinÂ³x) + 2(sin x) - 2(sinÂ³x) = 1 - 2(sinÂ²x)

(sin x) - 2(sinÂ³x) + 2(sin x) - 2(sinÂ³x)  + 2(sinÂ²x) - 1 = 0

-4(sinÂ³x) + 2(sinÂ²x) + 3(sin x) - 1 = 0

4(sinÂ³x) - 2(sinÂ²x) - 3(sin x) + 1 = 0


let z = (sin x)


4zÂ³ - 2zÂ² - 3z + 1 = 0


solve for z's then, solve for x's

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Algebra help - on calculus problem?

2 Answers