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Given that $\cos A=1/2$ and $\cos A$ and $\sin A$ have the same sign, find the value of $\sin(-A)$. If the question is referring to the first quadrant, where all trigonometric identities are positive, why is the value of $\sin (-A)$: $-\sqrt{3}/2$? Is it because this rule only applies to positive angles? Also is the rule $\sin(-A)= -\sin A $ applicable for every quadrant?

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    $\begingroup$ $\sin(-A)=-\sin(A$) for all $A$ (i.e., sin is an odd function) $\endgroup$ – J. W. Tanner Aug 21 '19 at 19:16
  • $\begingroup$ $A$ is in the first quadrant but $-A$ is not. $\endgroup$ – fleablood Aug 21 '19 at 19:40
  • $\begingroup$ Where is your problem? $\cos A$ and $\sin A$ have the same sin. But $\cos A$ and $\sin (-A)$ don't. $\endgroup$ – fleablood Aug 21 '19 at 19:41
  • $\begingroup$ If $\sin (-A) = -\sin A$ that means $\sin A$ and $\sin (-A)$ have opposite signs. SO if $\cos A$ and $\sin A$ have same signs. And $\sin A$ and $\sin (-A)$ have opposite signs. Then $\cos A$ and $\sin(-A)$ have opposite signs. So if $\cos A$ is positive, then $\sin(-A)$ is negative. $\endgroup$ – fleablood Aug 21 '19 at 19:52
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If the question is referring to the first quadrant, where all trigonometric identities are positive, ...

When you're told that the $\sin$ and $\cos$ of a quantity have equal sign, it doesn't necessarily mean they're both in the first quadrant (that happens only if you know both to be positive). What the statement means is that they're both either in the first, or else the third quadrant.

Is it because this rule only applies to positive angles?

It's not clear what rule you're talking about here. But most of the rules of trigonometry (i.e., identities) apply to any angle.

Also is the rule $\sin(-A)=-\sin A$ applicable for every quadrant?

Yes. This is just telling you that if you flip the unit circle along the real axis (so that all angles are mapped to their opposites), then the sines also change signs. Obviously, the cosines do not change, which is reflected in the rule $\cos(-A)=\cos A.$ These properties are known as oddness and evenness respectively.

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Remember the negative angle identities:

$$\sin(-A)=-\sin(A)$$ $$\cos(-A)=\cos(A)$$

To find the value of A, we have to solve the equation $\cos^{-1}(A)=\frac{1}{2}$ (Remember your 30-60-90 triangle)

Assuming that this is over the interval $[0, 360]$, you can get either $60^{\circ}$ or $300^{\circ}$ (Remember your reference angle formulas)

Because the question stated that the value of $\cos(A)$ and $\sin(-A)$ have the same sign, we will have to evaluate the $\sin(-300^{\circ})$ Why? Because only the cosine is positive in the fourth quadrant and because $\sin(-300) = -\sin(300) = --\frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{2}$

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Just do it.

$\cos A = \frac 12$

So what is $\sin A$? Well $\cos^2 A + \sin^2 A = 1$ and so $(\frac 12)^2 + \sin^2 A = 1$ so $\sin^2 A= 1 - \frac 14$ so $\sin A =\pm \frac{\sqrt {3}}2$.

But we are told $\sin A$ and $\cos A$ are the same sign so $\sin A =\frac {\sqrt 3}2$.

So what is $\sin -A$? Well, we have a rule that $\sin (-A) = -\sin A$ (and ,yes, that is true for all quadrants) so $\sin (-A) = -\sin A = -\frac {\sqrt 3}2$.

I'm not entirely sure where you trouble lies.

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Given that $\cos A=1/2$

Okay, that means that $\cos A > 0$ so $A$ is in the first or fourth quadrant.

and $\cos A$ and $\sin A$ have the same sign

Okay, that means that $A$ is in first or third.

In combo with $\cos A > 0$ that means $\sin A > 0$ and $A$ is in the first.

Is it because this rule only applies to positive angles?

What rule? There's nothing wrong and off here. We have $\sin A > 0$ and so $\sin (-A) = - \sin A < 0$. This is exactly what we'd expect.

Somehow I suspect you are reading on of those $-A$ as though it were positive $A$ but I don't know where.

Also is the rule $\sin(−A)=−\sin A$ applicable for every quadrant?

Yes, it is.

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