Maths - Please Help !!

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July 5th 2008, 04:18 PM
sweetG

Maths - Please Help !!
There are 4 questions that i need help in and if you can even answer one of them that would be great ! Plus the level of these questions are just highschool. [Im in my last year of highschool]. I'd truly appreciate the help :)

1) Prove this identity:

(cosecx + cotx)(cosecx - cotx) = (cotx)(tanx)

I started with LHS = (cosecx + cotx)(cosecx - cotx)
but don't know how to continue because these really confuse me.

2) Solve the following equation for theta (θ) over the given domain:

√3 sin 2θ = cos 2θ

Domain: 0 < θ < 360

3) Find all solutions in the interval -180 ≤ x ≤ 180 to the equation cosx + (sin^2)x = 5/4

4) The sides of a triangle are given by the lines:

2x - y + 3 = 0

x - 2y - 3 = 0
2x + y - 11 = 0
Find the area of this triangle.

I started off by saying for each equation:
when x = 0
y = ?
[I'd substitute x = 0 into every equation]

and I also did when y = 0, x = ? [by substituing y = 0 into all equations]
i got the points (-3/2, 1), (3, 0) and (11/2, 11)
Are the points that i got right ? And then i plotted the points which gave me a triangle. Now what do i do ?
July 5th 2008, 04:44 PM
badgerigar
Quote:

1) Prove this identity:

(cosecx + cotx)(cosecx - cotx) = (cotx)(tanx)

I started with LHS = (cosecx + cotx)(cosecx - cotx)
but don't know how to continue because these really confuse me.

you need to know the following

The most obvious way to do this is to just start simplifying both sides of the equation. Post again if you get stuck.

Quote:

2) Solve the following equation for theta (θ) over the given domain:

√3 sin 2θ = cos 2θ

Domain: 0 < θ < 360

the trick to this type of question is dividing by the cos part to get a tan, in this case

Quote:

3) Find all solutions in the interval -180 ≤ x ≤ 180 to the equation cosx + (sin^2)x = 5/4

Use
Quote:
4) The sides of a triangle are given by the lines:2x - y + 3 = 0
x - 2y - 3 = 0
2x + y - 11 = 0
Find the area of this triangle.

I started off by saying for each equation:
when x = 0
y = ?
[I'd substitute x = 0 into every equation]

and I also did when y = 0, x = ? [by substituing y = 0 into all equations]
i got the points (-3/2, 1), (3, 0) and (11/2, 11)
Are the points that i got right ? And then i plotted the points which gave me a triangle. Now what do i do ?
I'm not sure what you are doing here. To find the corners of the triangle you need to find the points where the lines intersect. You do this by taking each of the 3 possible pairs of lines and solving them simultaneously for x and y. To get the area, you probably are expected to use base X height/2. This is probably most easily done by finding the distances along each side and using Pythagoras' theorem to find the height.
July 5th 2008, 05:30 PM
galactus

One cool way you could go about it. Perhaps not the easiest. Using Heron's formula.

Find the intersection points of the lines and then use the distance formula to find the distance between them. Those will be the lengths of the sides of the triangle.

$2x+3=\frac{x}{2}-\frac{3}{2}$

$x=-3, \;\ y=3$

$\frac{x}{2}-\frac{3}{2}=-2x+11$

$x=5, \;\ y=1$

$2x+3=-2x+11$

$x=2, \;\ y=7$

$a=\sqrt{(-3-5)^{2}+(3-1)^{2}}=2\sqrt{17}$

$b=\sqrt{(-3-2)^{2}+(3-7)^{2}}=\sqrt{41}$

$c=\sqrt{(5-2)^{2}+(1-7)^{2}}=3\sqrt{5}$

Using Heron's formula, the semi-perimeter is

$s=\frac{(2\sqrt{17}+\sqrt{41}+3\sqrt{5})}{2}\appro x{10.68}$

Heron's formula to find area:

$\sqrt{s(s-a)(s-b)(s-c)}$
July 5th 2008, 07:35 PM
Soroban

He4llo, sweetG

Quote:

4) The sides of a triangle are given by the lines: . $\begin{array}{ccc}2x - y + 3 &= &0 \\ x - 2y - 3 &=& 0 \\ 2x + y - 11 &=& 0 \end{array}$
Find the area of this triangle.

Solve each equation for $y.$

. . $\begin{array}{cccc}y &=&2x + 3 & [1] \\ \\[-4mm] y &=&\frac{1}{2}x - \frac{3}{2} & [2] \\ \\[-4mm] y &=& \text{-}2x + 11 & [3]\end{array}$

Note that: . $[2] \perp [3]$ . . . It is a right triangle!

Find the intersections . . .

$[1] \cap [2]\!:\;\;2x + 3 \:=\:\frac{1}{2}x-\frac{3}{2} \quad\Rightarrow\quad x \:=\:\text{-}3,\;y\:=\:\text{-}3 \quad\Rightarrow\quad A(\text{-}3,\,\text{-}3)$

$[1] \cap [3]\!:\;\;2x+3 \:=\:-2x+11\quad\Rightarrow\quad x\:=\;2,\;y\:=\:7 \quad\Rightarrow\quad B(2,7)$

$[2] \cap [3]\!:\;\;\frac{1}{2}x-\frac{3}{2}\:=\:-2+11\quad\Rightarrow\quad x \:=\:5,\;y\:=\:1 \quad\Rightarrow\quad C(5,1)$

The right angle is at $C.$
Hence, $AC$ is the base, $BC$ is the height.

Go for it!