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[tt_deeb]

I'm having some trouble with these.

1.) Solve for the exact value for tan (arccosine (-1/4)?

2.) sin^2(2x) + 3 cos^2(2x) = 3/2

I need exact values for x. I understand you can substitute either/or but then I get lost in the algebra.

3.) I have 30ft of fencing and I want to build a triangle against my house. If I don't need to put fencing on the side of my house (the base?), what's my maximum area? What's the length of the base. (this is probably hard to do without a diagram - I have NO CLUE how you would begin this in order to find the max area).

4.) Solve for the exact value for arccos(sin(2005)?

Any help would be greatly appreciated.

 

[tt_deeb]

Guys, I really need these.

 

[nitewulf]

2)
sin^(2x) + 3[1-sin^(2x)] = 3/2 (recall sin^2 + cos^2 = 1)
sin^(2x) + 3 - 3sin^(2x) = 3/2
3 - 2sin^(2x) = 3/2
3 - 3/2 = 2sin^(2x)
3/2 = 2sin^(2x)
3/4 = sin^(2x)
sqrt(3)/2 = sin(2x) (square root of both sides)
arcsin(sqrt(3)/2) = 2x (arcsin of both sides)
2x = 60 deg
x = 30 deg = pi/6

 

[automagnus]

edi:// i suck at math NVM

 

[NohWun]

I'm having some trouble with these.

1.) Solve for the exact value for tan (arccosine (-1/4)?

Come on! Learn how to read that:

: What's the tan of the angle whose cosine is -1/4?

If the cosine of the angle is -1/4, what's that tell you?

What's the definition of cosine? It's the length of the "adjacent" side over the hypotenuse (the longest side).

So you can construct a triangle with sides 1 and 4 and then use Pythagorus's theorem to find the other side:

: 1*1 + b*b = 4*4
: b*b = 4*4 - 1*1
: b*b = 15
: b = sqrt(15)

So now you know the length of all 3 sides (keep in mind the "1" side has a negative sign associated with it). Now all you need to know the definition of tan, which is the length of the opposite side over the adjacent side. In other words,

: answer = sqrt(15)/-1
: answer = -sqrt(15)

 

[NohWun]

I'm having some trouble with these.
...
4.) Solve for the exact value for arccos(sin(2005))?

Any help would be greatly appreciated.

Once again:

: What's the angle who's cosine is sin(2005)?

In this case, we're starting with an angle, 2005.
If this is in degrees, it's really the same angle as 205 (subtract 5 multiples of 360).
Now, sin is the opposite over hypotenuse.
At this point, it's good to draw a diagram:

The angle 205 is represented by A in the figure.
The value of it's sin is represented by Y (next to the vertical side).

What we want to do is to take that Y and make it a cosine value instead.
That's done by drawing the horizontal side with length Y as shown.
It goes to the left, since the original Y is negative.
The angle associated with this line is labeled B.

The angle A is 25 degrees after the 180 mark (180 + 25 = 205).
The angle B is 25 degrees after the 90 mark, thus B = 90 + 25 = 115.

That's one way to solve the problem. Your calculator would have told you the same thing.

 

[NohWun]

I'm having some trouble with these.
...
3.) I have 30ft of fencing and I want to build a triangle against my house. If I don't need to put fencing on the side of my house (the base?), what's my maximum area? What's the length of the base. (this is probably hard to do without a diagram - I have NO CLUE how you would begin this in order to find the max area).
...

First you need a formula to express the area of a triangle:

: a = 1/2 * b * h

But there's too many variables, so now you impose the limits of the problem, which is the total length of the fence, but expressed in terms of the triangle's variables, which are b & h. This requires some trig, since h is not one of the sides.

Without any loss of generality, we can say that this is an isosceles triangle, and draw it with a vertical line going down the middle to represent "h". We can do this since the area is the same regardless of where the height line is; so let's just make it easy and put the height line in the middle. This lets us use Pythagorus's theorem and say that:

: (1/2 b)*(1/2 b) + h*h = s*s

We know that s + s = 30 (length of fencing), so in fact s=15. Substitute that in to the above formula:

: b*b/4 + h*h = 15*15
: h*h = 225 - b*b/4
: h = sqrt(225 - b*b/4)

Now substitute this expression for h into the original formula:

a = 1/2 b * sqrt(225 - b*b/4)

From here, we take the derivative and set it to 0 (this is how we find maximums and minimums).

a' = 1/2 b * 1/2 (225-b*b/4)^(-1/2) * (0-2b/4) + sqrt(225-b*b/4)*1/2
a' = -1/8 b*b * (225-b*b/4)^(-1/2) + sqrt(225-b*b/4)*1/2
a' = (-1/8 b*b + (225-b*b/4)*1/2 ) / sqrt(225 - b*b/4)
a' = (225/2 - b*b/4) / sqrt(225 - b*b/4)

So set this to 0, meaning we need to solve only:

0 = (225/2 - b*b/4)
b*b/4 = 225/2
b*b = 225*2
b = 15 * sqrt(2)

(You could also solve for the minimum of a^2, which is easier, and results in the same answer.)

There's an easier way to look at this problem, however.

You may have already solved the problem whereby you want to maximize the area of a rectangle with a fixed perimeter. The answer is when the sides are equal (the rectangle is a square). This is really the same problem, except we've split the square along its diagonal, then moved the pieces around to form a triangle. See here:

You'll notice that the base is equal to twice the height, which, if you solve for h above, is the same as the case above.