Just a little clarification...

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• Jan 13th 2010, 05:12 AM
Punch
Just a little clarification...
Is it true that to find the basic angle, we have to always use the positive number?

for example, $cosx = -1$
so to find the basic angle $= cos^-1 (1)$?
• Jan 13th 2010, 05:22 AM
dedust
Quote:

Originally Posted by Punch
Is it true that to find the basic angle, we have to always use the positive number?

for example, $cosx = -1$
so to find the basic angle $= cos^-1 (1)$?

what do you mean by basic angel?
if $\cos x = -1$, then $x = \cos^{-1} (-1)$
• Jan 13th 2010, 05:26 AM
Punch
Basic angle is the acute angle of a general angle
• Jan 13th 2010, 08:44 AM
Grandad
Hello Punch
Quote:

Originally Posted by Punch
Is it true that to find the basic angle, we have to always use the positive number?

for example, $cosx = -1$
so to find the basic angle $= cos^-1 (1)$?

Yes; because it's only acute angles that have their sine, cosine and tangent all positive.

Grandad
• Jan 14th 2010, 02:10 AM
Punch
How about this,

taking the cosine inverse of -1 will get me 180, that is obtuse. so i take the acute angle of it that is 0. And it is equals to cosine inverse of 1.

So this shows that whenever u sin cos tan something and get the obtuse angle, just use the acute angle? am i right?
• Jan 14th 2010, 02:51 AM
Grandad
Hello Punch
Quote:

Originally Posted by Punch
How about this,

taking the cosine inverse of -1 will get me 180, that is obtuse. so i take the acute angle of it that is 0. And it is equals to cosine inverse of 1.

So this shows that whenever u sin cos tan something and get the obtuse angle, just use the acute angle? am i right?

I think you need to be very careful to say exactly what you mean. There are lots of things to learn about trig functions, and you can't condense them into a single sentence! Let me try to clarify a bit:

• To find the angle $x$ between $0^o$ and $180^o$ if $\cos x$ is negative (say, $\cos x = -k$), find the acute angle whose cosine is $+k$ and subtract it from $180^o$.

• This works in exactly the same with with $\tan x$. If $\tan x$ is negative, find the angle whose tangent is the same positive number, and subtract from $180^o$.

• With $\sin x$, there are no negative values between $0^o$ and $180^o$: $\sin x$ and $\sin(180^o-x)$ have exactly the same positive value. You need angles outside this range if you want negative values of sine. For instance, if $\sin x = -k$, then you can find the acute angle whose sine is $+k$ and either add this to $180^o$ or subtract it from $360^o$.

I told you that, sadly, it wasn't as simple as you were trying to make it!

Grandad