We have that $\left|\cos\alpha\right|=\dfrac{3}{5}$ and $\alpha \in(90^\circ;180^\circ).$ Find $\sin\alpha-\cos\alpha.$

For every angle $\alpha \in(90^\circ;180^\circ)$ we have: $\sin\alpha\in(0;1), \cos\alpha\in(-1;0),\tan\alpha<0$ and $\cot\alpha<0.$ Can you give me a hint how to find $\sin\alpha-\cos\alpha$ without using the basic trig identity $\sin^2\alpha+\cos^2\alpha=1$ (we have proven it only for acute angles) or some other well-known identities (we have proven such only for acute angles). Thank you!

  • $\begingroup$ Can you use the fact that $\cos(180^\circ-x)=-\cos(x)$? $\endgroup$ – user170231 Dec 15 '20 at 16:19
  • $\begingroup$ No, we haven't studied it. $\endgroup$ – Katherine Dec 15 '20 at 16:20
  • $\begingroup$ What definition of $\sin \alpha$ are you working with ($\alpha > 90^\circ$)? $\endgroup$ – player3236 Dec 15 '20 at 16:25
  • $\begingroup$ use the unit circle, definition of cosine and Pythagorean theorem $\endgroup$ – Vasya Dec 15 '20 at 16:27

Since $\sin\alpha>0$, $\sin\alpha=\frac45$ so $\sin\alpha-\cos\alpha=\frac75$.

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    $\begingroup$ Thank you for the response! I am not sure I see why $\sin\alpha$ is equal to $\dfrac{4}{5}$. $\endgroup$ – Katherine Dec 15 '20 at 16:21
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    $\begingroup$ @LYI You have no choice but to use Pythagoras. What definitions of $\cos\alpha,\,\sin\alpha$ have you learned for obtuse $\alpha$? $\endgroup$ – J.G. Dec 15 '20 at 16:23
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    $\begingroup$ I don't see your point. How can I use Pythagoras when $\alpha$ is not an acute angle. I know that $\cos\alpha=\dfrac{Adjacent}{Hypotenuse}$ and $\sin\alpha=\dfrac{Opposite}{Hypotenuse}$. How to use it? $\endgroup$ – Katherine Dec 15 '20 at 16:27
  • $\begingroup$ From $|\cos\alpha|=\dfrac{3}{5}$ and $\cos\alpha\in(-1;0),$ we can conclude that $\cos\alpha=-\dfrac{3}{5},$ right? $\endgroup$ – Katherine Dec 15 '20 at 16:29
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    $\begingroup$ @LYI "How can I use Pythagoas" It depends on your answer to my question, "what definitions have you learned". For example, if you define trigonometry with a circle, Pythagoras is obvious for all angles. By contrast, if you've been told $\cos(180^\circ-\alpha)=-\cos\alpha,\,\sin(180^\circ-\alpha)=\sin\alpha$, you can reduce the problem to one in acute angles. $\endgroup$ – J.G. Dec 15 '20 at 16:30

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