# We have that $\left|\cos\alpha\right|=\frac{3}{5}$. What is $\sin\alpha-\cos\alpha$?

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!

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

## 1 Answer

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

• Thank you for the response! I am not sure I see why $\sin\alpha$ is equal to $\dfrac{4}{5}$. – Katherine Dec 15 '20 at 16:21
• @LYI You have no choice but to use Pythagoras. What definitions of $\cos\alpha,\,\sin\alpha$ have you learned for obtuse $\alpha$? – J.G. Dec 15 '20 at 16:23
• 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? – Katherine Dec 15 '20 at 16:27
• From $|\cos\alpha|=\dfrac{3}{5}$ and $\cos\alpha\in(-1;0),$ we can conclude that $\cos\alpha=-\dfrac{3}{5},$ right? – Katherine Dec 15 '20 at 16:29
• @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. – J.G. Dec 15 '20 at 16:30