Trigonometric Values (1 Viewer)

[FDownes]

I'm having probelms with another question, but it's a very simple one. The question is;

Use simpson's rule to find an approximation to _pi/8∫^pi/4 tan x dx, correct to 2 decimal places.

My only problem is that I'm not sure what interval to use. Could somone help me out here...?

 

[FDownes]

I'm still having trouble with the last question I asked. Could someone please help me out here?

 

[cwag]

I'm having probelms with another question, but it's a very simple one. The question is;

Use simpson's rule to find an approximation to _pi/8∫^pi/4 tan x dx, correct to 2 decimal places.

My only problem is that I'm not sure what interval to use. Could somone help me out here...?

how many values did they ask you to use

 

[Slidey]

Numerical approximations usually converge.

So take, say, 3 or 5 intervals. See what it gives for the first 2 decimal places. Then take more vintervals, and compare. If the decimal places haven't changed, go with that.

Alternatively you could interpret the question as "decide for yourself how many intervals to use".

 

[FDownes]

I forgot to mention, the question requires that you use 5 intervals. Sorry, that was probably the most important part and I left it out.

 

[cwag]

for five intervals... i find that this method seems to work...... subtract the smaller value, and divide by (n-1) where n is number of intervals to obtain the difference between then....

so ((1/4) - (1/8))/4
= 1/32
so your diffence should be 1/32

your table looks like this.....x | pi/8 | 5pi/32 | 3pi/16 | 7pi/32 | pi/4

you can go from there can u?

 

[FDownes]

Yep. Thanks.

 

[FDownes]

*sigh* Another one;

Evaluate _pi/12∫^pi/8 sec²2x dx, in exact form.

My working;

[tan 2x]_pi/12^pi/8
[tan 2(pi/8)] - [tan 2(pi/12)]
[tan pi/4] - [tan pi/6]
[1/√2] - [1/√3]


The answer is apparently (3 - √3)/6. Where am I going wrong?

 

[cwag]

*sigh* Another one;

Evaluate _pi/12∫^pi/8 sec²2x dx, in exact form.

My working;

[tan 2x]_pi/12^pi/8
[tan 2(pi/8)] - [tan 2(pi/12)]
[tan pi/4] - [tan pi/6]
[1/√2] - [1/√3]


The answer is apparently (3 - √3)/6. Where am I going wrong?

the integral of sec²2x is (tan 2x)/2

standard integrals sheet.... int: sec²ax = 1/a tan ax

 

[FDownes]

Damnit! I keep thinking questions are more complicated than they really are, and keep forgetting my basics. Thanks.

 

[tommykins]

Hahaha, I think it's just a lack of concnetration Fin.

Good luck on thursday

 

[FDownes]

Thanks, you too.

 

[FDownes]

One of the questions from my textbook has an answer that I'm not entirely sure about. The question is;

Find the derivative of tan x^o.

The textbook answer is pi/180 sec² x^o, but shouldn't the answer be 180/pi sec² x^o?

 

[tommykins]

x^o = pi/180, not 180/pi.

 

[FDownes]

Yeah... That's what I thought. But don't you flip the fraction when finding the derivative? Or am I getting confused?

 

[tommykins]

tan x^o

= tan (xpi/180)

dy/dx = pi/180 sec²(xpi/180) = pi/180sec²(x^o)

basically same as tan 2x

dy/dx = 2sec² 2x

Remember xpi/180, the pi/180 is a constant value.

 

[FDownes]

Okay, that makes sense. For some reason, I was convinced that was a special circumstance where the fraction was flipped.*sigh* I'm really bad at this topic. I need to do some serious revision.

While I'm here can anyone help me out with this question? It's;

Find the limit (as x approaches 0) of (1 - cos2x)/x².

 

[tommykins]

Haha I was just on that.

1-cos2x/x² = 1-(cos²-sin²)/cos² = sin²+cos²-cos²+sin²/x² = 2sin²/x²

Since x-> 0 and sin x -> x, sin²x = sinx.sinx = x²

Since its 2sin²x = 2x²/x² = 2.

 

[Slidey]

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

As x goes to zero, sin(x)/x goes to 1, thus the answer is 2.

 

[FDownes]

Thanks guys, I forgot about the special limit.