Evaluating sine of a shifted angle in quadrant II

Expert intro: This problem tests angle subtraction and quadrant sign rules. The key is to rewrite the angle so the reference angle is clear before applying sine identities.

Detailed walkthrough

Identify the angle position

We need to evaluate

$\sin\left(\frac{7\pi}{6}-\varepsilon\right)$

with the condition

$\varepsilon \in \left(\frac{\pi}{2},\pi\right).$

That means $\varepsilon$ is in Quadrant II, so $\sin\varepsilon>0$ and $\cos\varepsilon<0$.

Also,

$\frac{7\pi}{6}=\pi+\frac{\pi}{6}.$

So the expression is a shifted angle that can be handled with the sine difference identity.

Apply the sine subtraction formula

Use

$\sin(A-B)=\sin A\cos B-\cos A\sin B.$

Let $A=\frac{7\pi}{6}$ and $B=\varepsilon$.

Then

$\sin\left(\frac{7\pi}{6}-\varepsilon\right)=\sin\frac{7\pi}{6}\cos\varepsilon-\cos\frac{7\pi}{6}\sin\varepsilon.$

Now substitute the exact values

$\sin\frac{7\pi}{6}=-\frac12, \qquad \cos\frac{7\pi}{6}=-\frac{\sqrt3}{2}.$

So

$\sin\left(\frac{7\pi}{6}-\varepsilon\right)=\left(-\frac12\right)\cos\varepsilon-\left(-\frac{\sqrt3}{2}\right)\sin\varepsilon.$

Simplify the expression

This becomes

$\sin\left(\frac{7\pi}{6}-\varepsilon\right)=\frac{\sqrt3}{2}\sin\varepsilon-\frac12\cos\varepsilon.$

That is the exact simplified form.

A useful way to check the sign is to remember that in Quadrant II, $\sin\varepsilon$ is positive and $\cos\varepsilon$ is negative. Therefore the term $-\frac12\cos\varepsilon$ is positive, which fits the geometry of the shifted angle.

💡 Pitfall guide

A common mistake is to treat $\frac{7\pi}{6}$ as if it were in Quadrant I just because the formula looks familiar. It is actually in Quadrant III, so both sine and cosine are negative there. Another frequent error is reversing the subtraction identity into $\sin A\sin B-\cos A\cos B$, which is not correct. Finally, some students ignore the interval for $\varepsilon$ and try to assign a numeric value even though the problem only asks for an exact symbolic expression.

🔄 Real-world variant

If the problem changed to $\cos\left(\frac{7\pi}{6}-\varepsilon\right)$ with the same condition $\varepsilon\in\left(\frac{\pi}{2},\pi\right)$, you would use the cosine subtraction formula instead:

$\cos(A-B)=\cos A\cos B+\sin A\sin B.$

That would give

$\cos\left(\frac{7\pi}{6}-\varepsilon\right)=\left(-\frac{\sqrt3}{2}\right)\cos\varepsilon+\left(-\frac12\right)\sin\varepsilon.$

The method is the same: identify the quadrant, substitute exact unit-circle values, and simplify carefully.

🔍 Related terms

sine subtraction formula, unit circle quadrants, reference angle