Author Topic: HELP ME WE MATH!  (Read 801 times)

Kuiper7986

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HELP ME WE MATH!
« on: February 26, 2004, 06:08:18 am »
I\'m stumped guys I need help. I copied the questions down onto my PC then uploaded them, help me if anyone knows how to do this nonsense.

My name is NOT pronounced, \"Kway-per,\" it\'s pronounced \"Kye-per.\"

Rageburst

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« Reply #1 on: February 26, 2004, 06:47:07 am »
This is simple enough, but I forgot about csc, cot stuff.... useless to me in the end since I\'m into software stuff.

The first one goes like this:
You have to realize that log_x A = y means x^y=A.

We can work from the inner bracket (log_2 16)

This is basically solving 2^? = 16. Quite simply, 2^4 = 16 so now the problem reduces to log_4 4 = y

Easily, log_4 4 = 1 since 4^1 = 4. The answer is thus 1. Hope  you understood the typing.

For the third one,
just multiply the LHS by (1+cos x) / (1+cos x). This is basically one, so it doesn\'t change the value of the LHS.

When you do this, you\'ll get

LHS = (sin^2 x * sin^2 x cos x) / (1 - cos^2 x) ... but you know that 1 - cos^2 x = sin^2 x..... so the LHS equals cos x in the end.

As for the second one, try converting cot into csc using any  equation shown in book.
« Last Edit: February 26, 2004, 06:51:19 am by Rageburst »

Ghostslayer

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« Reply #2 on: February 26, 2004, 07:28:51 am »
For the 2nd one, you could always just draw a triangle.  Since cot B = 4/5, the side opposite the angle is length 5 and the side adjacent to the angle B is 4.

Solving for the hypotenuse using c^2 = a^2 + b^2 , you get the hypotenuse as sqrt(41).

csc of an angle is equal to the hypotenuse/opposite, so csc B would be -sqrt(41)/5. (See edit).

Although I haven\'t done that sorta stuff in a while... I\'m still trying to think where the resrtiction 180<=B<=270 comes into play.  If I think of it I\'ll let you know ;)

Edit: Thought of it already :P.   The restriction comes into play if it is positive or negative.  Using the CAST rule (dunno if you\'ve heard this or not... anyway 0-90 degress all is positive (the A) 90-180 only Sine is positive (S) 180-270 only tan is positive (T) and for 270-360 only cos is positive (C).)  Anyway, since the angle is between 180 and 270, thats why cot B is positive (since cot is 1/tan).  Therefore, the answer should be negative since csc (or 1/sin) is negative in that quadrant.
« Last Edit: February 26, 2004, 07:33:07 am by Ghostslayer »
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lynx_lupo

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« Reply #3 on: February 26, 2004, 09:42:44 am »
Just a typo kuiper, rageburst, in the end it\'s 1+cos(x).
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Kuiper7986

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« Reply #4 on: February 26, 2004, 04:27:37 pm »
thanks buds you just saved my life.
My name is NOT pronounced, \"Kway-per,\" it\'s pronounced \"Kye-per.\"