General Solution for Negative Cosine Ratio (1 Viewer)

[Starcraftmazter]

Hey guys.

Have my 11 prelim test tomorrow in ext, been studing for a few days. While going over my ext trig notes just now, I saw something a bit weird.

Question: Find the General Solution for Cos A = -1/sq(2)
Take note of the negative number.

Now the answer in my book for this question was n.360+-45. Yet it was marked with a cross, and the "correct" solution was written down as +-45+-(2n)^-1*180, which to me doesn't look very right, but could well be right, because I'm not that smart 8)

So if someone could confirm which is the right answer, or if they're both wrong, before 1am or so, that'd be great, thanks

 

[fishy89sg]

dont get your hopes up but

there were no general solutions in our 3 unit exam

 

[pLuvia]

The general solution to cosA is:
A=2npi+@
So,
cosA=-1/sqrt{2}
=cos(-45)
=cos(135)
A=n.360+135

 

[haque]

Cos is negative in the second and third quadrants, so if CosA=-1/sqrt2 then A=(2n-1)pi+-(pi/4)=(2n-1)180+or-45 (note that (2n-1 is always odd-this is significant for this result to be true). the negative of the expression is also true as cos(-A)=cosA

 

[Starcraftmazter]

Alright thanks guys.

I did some testing, and found that +-45+-(2n)^-1*180 works as a solution.
The troubling thing is, pLuvia's solution works just as well, except it's a million times more simple.

pLuvia, could you please explain how you got cos 135 from -cos 45, that would be great.

haque, your solution seems to give alternating positive/negative values of 1/sq{2} at each increasement of n.

also it might be worthy to mention, I'm not sure what you mean by pi in both your solutions.

And fishy89sg....can you see the future? Because if you can, I'd love to know how 8)

 

[Riviet]

Pi as in the top right one in this image:
http://upload.wikimedia.org/wikipedia/commons/7/7d/Greek_letter_pi.png

There is an identity involving cos^-1x:
cos^-1(-x)=pi - cos^-1x

.'. A = 2.pi.n + cos^-1(-1/sqrt2) = 2.pi.n +{pi - cos^-1(1/sqrt2)}
=2.pi.n +(pi - pi/4)
= 2.pi.n + 3pi/4, where 3pi/4 radians is equivalent to 135 degrees.

 

[haque]

u must be doing something wrong when u sub in values of n as it should not alternate. consider n=1 then we get A=180 +-45=135,225 for which if u take cos are negative 1/sqrt2. if n=2 then u get A=3times 180+-45 which also gives u 1 divided by negative sqrt 2. the 2n-1 is infront of the 180 only.

 

[Starcraftmazter]

Pi as in the top right one in this image:
http://upload.wikimedia.org/wikipedia/commons/7/7d/Greek_letter_pi.png

There is an identity involving cos^-1x:
cos^-1(-x)=pi - cos^-1x

I feared that....well....we haven't done that yet. And now thats another thing, to not look forward to lol.

u must be doing something wrong when u sub in values of n

Appologies haque - my mistake. I'm pretty sure I forgot to multiply my values of n by 2, before subtracting the 1 and going on the with process.

That'll teach me not to do maths at night.....aaa who am I kidding, no it won't 8)

 

[haque]

No worries-i was surprised because i gave my answer in a form similar to what ur textbook wanted. by convention most of the angles given that are either subtracted from 180 degrees or subtracted from 360 degrees are usually given with an acute angle-so the textbook stated 45 rather than 135-however both are right.

 

[Starcraftmazter]

We don't use textbooks :s

But the main thing is, we haven't been tought these pi identities everyone seems to use, I assume thats year 12 stuff.

 

[pLuvia]

It makes the answer much nicer than having just numbers, I learnt it in year 11 near the end. Pi is denoted as 180 degrees, pi/4 is 45 degrees and so on. You'll learn it very soon

 

[Starcraftmazter]

Alrighty.

 

[SoulSearcher]

Learnt as part of the trig functions topic.