Trigonometric Values (1 Viewer)

[SpinCobra]

I have a feeling I'm way off. Is it:

Root(3)/32?

 

[FDownes]

Bingo! Could I have a look at your working please? I'm really stuck on this one.

 

[Mark576]

I got the same answer. So it's likely that we're correct.

 

[SpinCobra]

₀∫^4√3 dt / (16 + t²)^3/2

-----------------------
Let..
t = 4 tan Ө
dt = 4 sec²Ө dӨ

when t = 0, Ө = 0
when t = 4√3, Ө = pi/3

-----------------------

Therefore

₀∫^4√3 dt / (16 + t²)^3/2

= ₀∫^pi/3 (4 sec²Ө) dӨ / (16 + 16 tan²Ө)^3/2

= ₀∫^pi/3 (4 sec²Ө) dӨ / (4√(1 + tan²Ө))³

= ₀∫^pi/3 (4 sec²Ө) dӨ / (4√(sec²Ө))³

[√(sec²Ө) = sec Ө]

= ₀∫^pi/3 (4 sec²Ө) dӨ / (4³ sec³Ө)

(Cancel crap on top and bottom?)

= ₀∫^pi/3 dӨ / 16 sec Ө

(Take 1/16 outside integral)

= 1/16₀∫^pi/3 dӨ / sec Ө

= 1/16₀∫^pi/3 cos Ө dӨ

Integrate.

 

[SpinCobra]

I wish I had a scanner. Typing this shit up takes forever.

 

[Slidey]

It's nice of you to try and do it neatly man, but seriously just use sqrt, ^, Int, ^(1/2), @ for theta, etc.

We'll still understand you.

 

[FDownes]

You guys have been awesome so far, lets see if we can keep up the record with one more problem;

a) Write down an expression for sin 3x in terms of sin x.

b) Hence find ₀∫^pi/6 sin³x dx.

 

[tommykins]

From memory..

sin 3x = 4sinx - 3sin³x
3sin³x = 4sinx - sin 3x
sin³x = (4sinx - sin3x)/3
sin3x dx.

= 1/3 ∫ 4sinx - sin3x dx
= 1/3[ -4cosx + cos3x/3 ]0 -> pi/6

Go on from there -

PS. I feel really iffy on the trig expansaion, I did it in clas and couldn't find the shet so I'm doing ti solely based on memory, if I'm wrong please point it out.

 

[FDownes]

Sorry I'm a bit late, but thanks.

 

[cwag]

Edit. sin3x in terms of sin x = 3sin x- 4 sin³x
Proof...sin 3x = sin(2x+x)
=2sinxcos²x + sin x - 2sin³x
=2sinx - 2 sin³x + sinx - 2sin³x
=3sinx - 4sin³x

therefore ₀∫^pi/6 sin³x
=1/4 ₀∫^pi/6 3sinx-sin3x
=1/4 [-3cosx + (cos 3x)/3]₀^pi/6
=1/4 ((-3root3)/2 - (-3+1/3))
=1/4 ((-3root3)/2 + 8/3)
approx = 0.017

 

[FDownes]

Stuck once again. Here we go;

Given that y = e^-x(cos2x + sin2x), show that y'' + 2y' + 5y = 0

 

[Zephyrio]

Stuck once again. Here we go;

Given that y = e^-x(cos2x + sin2x), show that y'' + 2y' + 5y = 0

Differentiate using the product rule.

Find the second derivative.

Then sub these values into y'' + 2y' + 5y. Should equal 0.

 

[FDownes]

My problem is that I'm having some trouble differentiating the function. Could somebody walk me through the problem, please?

 

[cwag]

y = e^-x (cos 2x + sin 2x)

y'=u'v + v'u (product rule)

u'v = -e^-x(cos2x+sin2x)
v'u = (-2sin2x + 2cos2x)e^-x.......because derivative of cos ax = -asinax, and derivative of sin ax = acos ax

therefore y'= -e^-x(cos2x+sin2x) + (-2sin2x + 2cos2x)e^-x

if we take a common factor of e^-x
we get y' = e^-x{(-cos2x-sin2x)+(-2sin2x + 2cos2x)}

by simplifing further we get
y'= e^-x(cos2x - 3sin2x)

Edit: now y" = -e^-x(cos 2x - 3 sin2x)+e^-x(-2sin2x-6cos 2x)
y"= e^-x{(-cos 2x + 3 sin2x) + (-2sin2x - 6cos2x)}
y"= e^-x(-7cos2x + sin2x)

therefore. y" + 2y' + 5y = e^-x(-7cos2x + sin2x) +2e^-x(cos2x - 3sin2x) + 5e^-x (cos 2x + sin 2x)

= -7cos(2x).e^-x+ sin(2x).e^-x + 2cos(2x).e^-x - 6sin(2x)e^-x + 5cos(2x)e^-x + 5sin(2x)e^-x

which = 0 hooray.....please don't tell me theres an easier way to do that thou

 

[Mark576]

y=e^-x(cos2x+sin2x)
ye^x=cos2x+sin2x
ye^x+y'e^x=-2sin2x+2cos2x [differentiating both sides]
e^x(y+y')=-2sin2x+2cos2x
e^x(y+y')+e^x(y'+y'')=-4cos2x-4sin2x [differentiating both sides]
e^x(y''+2y'+y)=-4(cos2x+sin2x)
y''+2y'+y=-4y
y''+2y'+5y=0

Quite a neat method.

 

[cwag]

y=e^-x(cos2x+sin2x)
ye^x=cos2x+sin2x
ye^x+y'e^x=-2sin2x+2cos2x [differentiating both sides]
e^x(y+y')=-2sin2x+2cos2x
e^x(y+y')+e^x(y'+y'')=-4cos2x-4sin2x [differentiating both sides]
e^x(y''+2y'+y)=-4(cos2x+sin2x)
y''+2y'+y=-4y
y''+2y'+5y=0

Quite a neat method.

haha...looks like u got me again mark

 

[Mark576]

haha...looks like u got me again mark

hehe

 

[FDownes]

Thanks very much...

Oh, and one quick question while I'm here; would cos²2x be equal to 1/2(1 - cos4x)?

 

[cwag]

Thanks very much...

Oh, and one quick question while I'm here; would cos²2x be equal to 1/2(1 - cos4x)?

No, close...its 1/2 (1 + cos4x)
its always double the angle and a - sign for sin and + sin for cos...eg....cos²5x = 1/2(1 + cos10x).....sin²x/2 = 1/2(1-cosx)

 

[FDownes]

Ah, that's my broblem. Thanks for clarifying that.