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 $\displaystyle = cos^-1 (1)$?

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- Jan 13th 2010, 04:12 AMPunchJust 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 AMdedust
- Jan 13th 2010, 04:26 AMPunch
Basic angle is the acute angle of a general angle

- Jan 13th 2010, 07:44 AMGrandad
- Jan 14th 2010, 01:10 AMPunch
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 AMGrandad
Hello PunchI 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