Trigonometric Values (2 Viewers)

[FDownes]

Hmm.. My calculations for this question seem to end up coming out all wrong. Could someone please show me theirs so I can see what I'm doing incorrectly? The question is;

Use the trapezoidal rule with 4 subintervals to find, correct to 3 decimal places, an approximation to the volume of the solid formed by rotating the curve y = sin x about the x-axis from x = 0.2 to x = 0.6

 

[FDownes]

I skipped the last question, but now I have a new problem;

Show that sinx cosx = 1/2 sin2x, and hence or otherwise find the exact value of ₀∫^pi/8 sin²x cos²x.

The first half of this question is easy, it's the second half that's giving me probelms. Could someone please show me their working?

 

[Mark576]

sinx.cosx = 1/2sin2x
sin²x.cos²x = 1/4sin²2x
sin²2x = 1/2 - 1/2cos4x

That should help you.

 

[FDownes]

Ah, I see what my problem was. I was assuming sin²x.cos²x = 1/2sin²2x, not sin²x.cos²x = 1/4sin²2x.

Thank ye kindly good sir.

EDIT: Ack! I'm so close to getting the answer, but by my working I end up with (pi - 4)/64 where it should be (pi - 2)/64, at least, according to the textbook...

Here's my working;

₀∫^pi/8 1/4 sin²2x
1/4 ₀∫^pi/8 sin²2x
1/4 [x/2 - 1/4sin4x]₀^pi/8
1/4 [(pi/8)/2 - 1/4sin4(pi/8) - (0)/2 + 1/4sin4(0)]
1/4 [pi/16 - 1/4]
pi/64 - 1/16
(pi - 4)/64

 

[Mark576]

Ah, I see what my problem was. I was assuming sin²x.cos²x = 1/2sin²2x, not sin²x.cos²x = 1/4sin²2x.

Thank ye kindly good sir.

EDIT: Ack! I'm so close to getting the answer, but by my working I end up with (pi - 4)/64 where it should be (pi - 2)/64, at least, according to the textbook...

Here's my working;

₀∫^pi/8 1/4 sin²2x
1/4 ₀∫^pi/8 sin²2x
1/4 [x/2 - 1/4sin4x]₀^pi/8
1/4 [(pi/8)/2 - 1/8sin4(pi/8) - (0)/2 + 1/4sin4(0)]
1/4 [pi/16 - 1/8]
pi/64 - 1/32
(pi - 2)/64

Fixed.

 

[FDownes]

Thanks again.