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Topic review (newest first)

2005-08-26 23:32:33

Prove that (sec^2 x + 2 tan x)/ (1 - tan^2 x) = (cos x + sin x)/(cos x - sin x)
Hence solve the equation (sin x +cos x)/(sin x - cos x) = tan x for 0 < x < 2pi.

2005-08-26 22:09:31

cosecA + cotA = 3
To find cosecA - cotA,
one should remember,
cosecA - cotA = 1
From this, we get
(cosecA+ cotA)(cosecA - cotA)=1
Given cosecA + cotA = 3,

cosecA + cotA + cosecA-cotA = 3+1/3 = 4/3
2cosecA = 4/3
sinA=3/2 funny smile

2005-08-26 20:41:34

thanks ajp3. But I got another question on trigo...

Given cosec A + cot A = 3

and thus evalute cosec A -cot A

and find cos A

2005-08-26 11:31:05

ahgua, do not worry about dividing by cos x in the (correct) sol'n given by wcy; sin x/cos x is defined as tan x, so we do not need to worry about division by zero, the dfn of tan x already forbids it!

another way to show x=0 is a sol'n was already given by wcy in that soln; the sin x = 0 half of the soln...

2005-08-24 22:16:35

You are right!
When x=0,
2sinx.cosx + cos^2x = 1
But how do we get a solution x=0?
2sinx.cosx + cos^2x = 1
sin2x + cos^2x = 1,
sin2x = 1-cos^2x
sin2x = sin^2x
this would be true only when x=0 smile
But is there any (other)mathematical way it can be shown that x=0 is a solution?

2005-08-24 20:59:24

wcy wrote:

sin x=2 cos x
tan x=2

Are you sure we can divide sin x by cos x?
what if cos x = 0 ...
then, the answer will be undefined!

2005-08-24 20:41:28

1=cos^2 x+sin^2 x
then, group all the sin terms...

2 sinx cosx +cos^2x=cos^2x +sin^2 x
2 sinx cosx=sin^2 x
sin^2 x-2 sinx cos x=0
sin x(sin x-2 cos x)=0
sin x=0 or sin x=2 cos x
or tan x=2

2005-08-24 20:14:00

Solve this:

2sinx.cosx + cos^2x = 1

2005-08-23 22:24:41

Making a right-angled triangle out of AB, the line going down from B and the line going right from A, we have one angle and one length. The 'opposite' side is 0.8m, the angle is 60 and we want to find the 'adjacent' side. tanθ=o/a, so tan 60=0.8/a, so a=0.8/tan 60=0.461... There are 21 of these, so 9.6995... of the 10m is being taken up by them. That leaves 0.3005... for CD. There are 2 of CD, so they will get 0.1503... each.

Making another right-angled triangle out of CD and the corner, we now know the side length and we want to know the hypotenuse. cosθ=a/h, so cos 60=0.1503.../h, so h=0.1503.../cos 60=0.30m to the nearest cm, which is length CD.

The outer frame has two lengths and two widths on it, so in total it is 10+10+0.8+0.8=21.6m long and one length AB is 0.8/sin 60=0.9238...m, so 21 of them are 19.40m to the nearest cm.

2005-08-23 21:44:53

The structure below measures 10m in length by 0.8m in width. In total, there are 21 diagonal pieces exactly like AB and two extra pieces- CD. Pieces CD are parallel to pieces AB. Pieces AB are inclined at 60 degrees from the horizontal.

Please help mefind the length (in millimetres) of piece CD as i already know how to find the length of  AB.

After that i need to find the total length of the out frame and all 21 pieces of AB.

Any help is greatly appreciated.

Note: the vertical lines (apart from those on the outer frame) are not a part of the actual structure. They are merely usaed to form triangles so certain lengths can be calculated.


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