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The question is :

Let $\cos\theta= \frac{-1}{4}$ and $\tan \theta> 0$. Find the remaining trigonometric functions.

I got

$\sin \theta = -\frac {\sqrt{15}}{4}$

$\tan \theta = \sqrt {15}$

$\csc \theta= \frac{-4\sqrt{15}}{15}$

$\sec \theta= -4$

$\cot \theta= \frac{\sqrt{15}}{15}$

Is this correct?

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    That many questions asked and still such poor formatting. It's high time you learn [MathJax](http://meta.math.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference).2017-02-06
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    I think all these correct.2017-02-06
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    Sine must be negative...which fits what you wrote about cosecant...2017-02-06
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    @ShraddheyaShendre Please do pay attention when you edit: you ommited a minus sign in $\;\sin\theta\;$ ...2017-02-06
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    @DonAntonio - I am not sure whether it was a minus sign or a hyphen. But seeing the negative cosec, I now think that it indeed was a minus sign. Sorry.2017-02-06
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    Also, I guess now only the OP can edit the question (since the edit is small). Or else, some redundant edits can be done to incorporate the change to correct the mistake I introduced.2017-02-06
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    Once you have both cos and sin, the rest follow from them.2017-02-07

1 Answers 1

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No. They all are correct except $Sin\hspace{1mm}\theta$.
From the information given in question, we are certain that the quadrant talked about is 3rd. Since $Tan\hspace{1mm}\theta>0$ and $Cos\hspace{1mm}\theta=-\frac{1}{4}$
So we can use $Sin^2\hspace{1mm}\theta+Cos^2\hspace{1mm}\theta=1$ and $Sin\hspace{1mm}\theta<0$ in 3rd quadrant.
So $Sin\hspace{1mm}\theta$ is negative.
And then, everything else is straightforward. I hope this helps.

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    Then no: not all are correct since, as you say, we're in the third quadrant and thus sine **must be negative** ...2017-02-06
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    Wow. I don't know how I didn't notice that before. Yes, sine will be negative. Thanks for pointing that out!2017-02-06
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    @DonAntonio In main post there was a **negative** back of sine, in edit it changed.2017-02-06
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    @MyGlasses True indeed, I just checked.2017-02-06