# Just a little clarification...

Printable View

• Jan 13th 2010, 04:12 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, $\displaystyle cosx = -1$
so to find the basic angle $\displaystyle = cos^-1 (1)$?
• Jan 13th 2010, 04:22 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, $\displaystyle cosx = -1$
so to find the basic angle $\displaystyle = cos^-1 (1)$?

what do you mean by basic angel?
if $\displaystyle \cos x = -1$, then $\displaystyle x = \cos^{-1} (-1)$
• Jan 13th 2010, 04:26 AM
Punch
Basic angle is the acute angle of a general angle
• Jan 13th 2010, 07:44 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, $\displaystyle cosx = -1$
so to find the basic angle $\displaystyle = cos^-1 (1)$?

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

Grandad
• Jan 14th 2010, 01:10 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, 01:51 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 $\displaystyle x$ between $\displaystyle 0^o$ and $\displaystyle 180^o$ if $\displaystyle \cos x$ is negative (say, $\displaystyle \cos x = -k$), find the acute angle whose cosine is $\displaystyle +k$ and subtract it from $\displaystyle 180^o$.

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

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

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

Grandad