Trigonometric ratios involving negative angles

Question

Given that $\cos \theta = \dfrac{3}{5}$ and csc is positive

a) which quadrant is $\theta$ in? Hence deduce the quadrant that $-\theta$ is in.

so for question it is 2nd quadrant

b) Without finding the angle $\theta$, obtain the exact values of

(i) $\cos (-\theta)$

(ii) $\sin (-\theta)$

(iii) $\tan (-\theta)$

For part B i'm confused to what formula should be use?

Answer 1

(a) Since $\cos\theta=3/5>0$ and $\csc\theta=1/sin\theta>0\implies\sin\theta>0$. Thus, $\theta$ is in the first quadrant or $0<\theta<90\implies -90<-\theta<0$, and hence, $-\theta$ is in the fourth quadrant.

(b) Since cosine function is an even function and sine and tangent function are both odd functions, then

$\cos(-\theta)=\cos\theta$;

$\sin(-\theta)=-\sin\theta$;

$\tan(-\theta)=-\tan\theta$;

and so, using phytagorean theorem we have $\sin\theta=\sqrt{1-\cos^2\theta}=\sqrt{1-9/25}=4/5$. And thus

$\cos(-\theta)=\cos\theta=3/5$;

$\sin(-\theta)=-\sin\theta=-4/5$;

$\tan(-\theta)=-\tan\theta=-\sin\theta/\cos\theta=-(4/5)/(3/5)=-4/3.$;

Answer 2

As $\displaystyle\cos\theta>0,\sin\theta=\frac1{\csc\theta}>0,\theta$ lies in the first Quadrant, hence $\displaystyle\tan\theta>0$

Hence, $-\theta$ lies in the fourth Quadrant where cosine ratio is positive unlike the rest two

So here, $\displaystyle\cos(-\theta)=\cos\theta$ and $\displaystyle\sin(-\theta)=-\sin\theta$ and so on

These can be derived easily using subtraction formulae $\displaystyle\sin(A-B)=\sin A\cos B-\cos A\sin B$ etc.