Exact Ratios? (1 Viewer)

[enak101]

I need to catch up on some prelim topics and I'm kind of getting stuck on exact ratios

I need to find the exact ratio of cosec(45)

I get 1/sin(45)

Which is 1/root 2/2

So you get the reciprocal which is 1*2/root 2

Which is 2/root 2, rationalize the denominator by multiplying both the numerator and denominator by root 2

That makes 2 root 2/2

The answer in my textbook is just root 2, any idea how to do these? Also, general help with adding and working with exact ratios would be nice.

 

[SpiralFlex]

That's the long way, or you could consider sin 45 to be 1 over square root 2,

 

[SpiralFlex]

Note: The second way is quicker if the adjacent numbers next to the viculum are equal. You could reciprocate it in your head easily.

 

[enak101]

Ah I see now, the 2's cancel out in the fraction lol, knew it would be something like that. I do like the second way heh, obviously that isn't going to work all the time but a good way of doing it. Thanks heaps for the help.

 

[SpiralFlex]

Ah I see now, the 2's cancel out in the fraction lol, knew it would be something like that. I do like the second way heh, obviously that isn't going to work all the time but a good way of doing it. Thanks heaps for the help.

It will work all the time since square root 2/ 2 is equivalent to 1/ square root 2.

If something is equivalent to something then it will always work, (The transitive property)

If A=B, B=C, then A=C. But if Team A defeats Team B, Team B defeats Team C. Does that mean Team A will defeat Team C?

 

[enak101]

Oh I see how your second method works now. If Team A defeats Team B and Team B defeats Team C, that doesn't necessarily mean Team A would beat C, if you are talking about sports lol, but I know what you are talking about with A=B B=C A=C

 

[SpiralFlex]

Oh I see how your second method works now. If Team A defeats Team B and Team B defeats Team C, that doesn't necessarily mean Team A would beat C, if you are talking about sports lol, but I know what you are talking about with A=B B=C A=C

Correct young grasshopper.

 

[enak101]

I feel dumb now.

I can't get cot 30+ cot 60

1/tan 30 + 1/tan 60

= 1/1/root 3 + 1/root 3
= root 3/1 + 1/root 3

Rationalize denominator of second one

root 3/1 + 1+root 3/3

=2 root 3/4

Alas, the answer is 4 root 3 over 3, can't see how they got that, I've been getting most of the questions right but there are a few that just escape me lol.

 

[someth1ng]

Just square it in your calculator and you'll be able to see...

 

[Timske]

<a href="http://www.codecogs.com/eqnedit.php?latex=\\cot30@plus;cot60\\\\ = \frac{1}{tan30}@plus;\frac{1}{tan60} \\= \frac{1}{\frac{1}{\sqrt{3}}}@plus;\frac{1}{\sqrt{3}} \\=\sqrt{3}@plus;\frac{1}{\sqrt{3}}\\=\frac{4}{\sqrt{3}} \\\frac{4}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3}" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\\cot30+cot60\\\\ = \frac{1}{tan30}+\frac{1}{tan60} \\= \frac{1}{\frac{1}{\sqrt{3}}}+\frac{1}{\sqrt{3}} \\=\sqrt{3}+\frac{1}{\sqrt{3}}\\=\frac{4}{\sqrt{3}} \\\frac{4}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3}" title="\\cot30+cot60\\\\ = \frac{1}{tan30}+\frac{1}{tan60} \\= \frac{1}{\frac{1}{\sqrt{3}}}+\frac{1}{\sqrt{3}} \\=\sqrt{3}+\frac{1}{\sqrt{3}}\\=\frac{4}{\sqrt{3}} \\\frac{4}{\sqrt{3}}*\frac{\sqrt{3}}{\sqrt{3}} = \frac{4\sqrt{3}}{3}" /></a>

 

[SpiralFlex]

I feel dumb now.

I can't get cot 30+ cot 60

1/tan 30 + 1/tan 60

= 1/1/root 3 + 1/root 3
= root 3/1 + 1/root 3

Rationalize denominator of second one

root 3/1 + 1+root 3/3

=2 root 3/4

Alas, the answer is 4 root 3 over 3, can't see how they got that, I've been getting most of the questions right but there are a few that just escape me lol.

 

[enak101]

Thanks, I keep forgetting the 1/whatever just makes it that root. Thanks again.

 

[SpiralFlex]

Thanks, I keep forgetting the 1/whatever just makes it that root. Thanks again.

No problem just keep practicing.

 

[enak101]

Lol, turns out I'm way better at finding exact values of pronumerals in triangles then just the exact values of different ratios. Guess I'll just have to ask teacher about those specifically. Thanks for the help, I've at least got a bit more of an idea now.