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Given $\tan A = \sqrt3$ and $\csc B = -\sqrt2$ where $A,B \in[\pi,\frac{3}{2}\pi]$,

evaluate

$\cos^2A + (\cot B)(\sin A)$.

I haven't been able to figure out the angle for $B$.

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  • $\begingroup$ Edited your question to insert latex. I hope i didn't change your question in any way. $\endgroup$ – sxd Nov 22 '11 at 4:19
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    $\begingroup$ You don't need to figure out what angle $A$ and angle $B$ are. But in fact, angle $B$ is not hard. Its cosecant is $-\sqrt{2}$, so its sine is $-1/\sqrt{2}$. Familiar? In your intervaal, it is hafway between $\pi$ and $3\pi/2$. $\endgroup$ – André Nicolas Nov 22 '11 at 4:20
  • $\begingroup$ I have also done some editing. I hope the question remains the same. $\endgroup$ – smanoos Nov 22 '11 at 4:20
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$$\csc B = -\sqrt 2 \implies \frac 1 {\sin B}=-\sqrt 2\implies\frac {-1} {\sqrt 2}=\sin B$$ Certainly now you can figure out what $B$ is.

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  • $\begingroup$ yeah i got up to there then i did sin -1(-1/root2) one thing im not sure would i use the angle i get or add PI to it $\endgroup$ – Faraz Nov 22 '11 at 4:25
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    $\begingroup$ @Faraz Well, you can't add $\pi$ to it, since sine does not have period $\pi$. You can add $2\pi$ to it though, which will put it into the interval you specified at the beginning. $\endgroup$ – process91 Nov 22 '11 at 4:28
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Using the identity $1+\tan^2A=\sec^2A$, you have that $\sec^2A=4$. i.e $\cos^2A=\frac{1}{4}$. So $\cos A=-\frac{1}{2}$, since $A$ is in the third quadrant. From this we get $\sin A=-\frac{\sqrt3}{2}$, again since $A$ is in the third quadrant.

Now, using the identity $1+\cot^2B=\csc^2B$, you get $\cot B=1$. This is positive since $\tan$ is positive in the third quadrant (which means $\cot$ is also positive), and so the result follows.

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Hints: We can find an explicit expression without knowing the angles (the angles are, however, not hard to find).

For example, recall the identity $1+\tan^2 A=\sec^2 A$. If you know $\tan A$, then you know $\sec^2 A$, and therefore you know $\cos^2 A$.

If you know $\cos^2 A$, you know $\sin^2 A$. Therefore you almost know $\sin A$, apart from sign. But since $A$ is in the third quadrant, you know the sign of $\sin A$.

You know $\csc B$, and want to know $\cot B$. There are various approaches. For example, from $\sin^2x+\cos^2 x=1$, dividing both sides by $\sin^2 x$, you get $1+\cot^2 x=\csc^2 x$. If you know $\csc B$, you can find $\cot^2 B$, and then, $\cot B$. Again, you have to worry about sign. But that's not hard, $\tan$ is positive in the third quadrant.

Or else you know $\csc B$, so you know $\sin^2 B$, so you know $\cos^2 B$, so you know $\cot^2 B$.

Comment: We know that $\csc B=-\sqrt{2}$. Thus $\sin B=-1/\sqrt{2}$, a familiar quantity. We know that the sine of $\pi/4$ is $1/\sqrt{2}$. So halfway between $\pi$ and $3\pi/2$, the sine is $-1/\sqrt{2}$.

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  • $\begingroup$ cos(4/3Pi)^2 + (cot5/4Pi)(sin5/4Pi) = =(-1/2)^2 + (1 )(root2/2) = (1/4) + (root2/2) = (1+2root2)/4 I didint realize the fact that 1/root2 =root2/2 ( i memorized root2/2 as part of the unit circle) (don't have enough rep yet to post my own answers) $\endgroup$ – Faraz Nov 22 '11 at 4:48
  • $\begingroup$ @Faraz: Your question say $\cot B \sin A$. Note that $\sin A=-\sqrt{3}/2$, so there is a slip in your answer. $\endgroup$ – André Nicolas Nov 22 '11 at 5:09

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