my working out was this what was wrong?
sin 240 = sin(270-30)
=-sin(30)
=1/2

Your working out is almost correct, but there's a slight sign error in the final step. Here's a breakdown of where the mistake occurred:

You correctly identified:
\[
\sin(240^\circ) = \sin(270^\circ - 30^\circ)
\]
This step is fine because you're expressing \(240^\circ\) as \(270^\circ - 30^\circ\).

Next, the identity \(\sin(270^\circ - \theta) = -\cos(\theta)\) is what applies here. However, you seem to have mistakenly used the sine function instead of cosine. So the correct identity is:
\[
\sin(270^\circ - 30^\circ) = -\cos(30^\circ)
\]
We know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\), so:
\[
\sin(240^\circ) = -\cos(30^\circ) = -\frac{\sqrt{3}}{2}
\]

The error occurred in the last step where you wrote \( \sin(30^\circ) = \frac{1}{2} \), but you should have used \(-\cos(30^\circ) = -\frac{\sqrt{3}}{2}\) instead.

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Would you like me to explain this further or solve any other similar problems?

Here are 5 related questions:
1. Can you explain the difference between sine and cosine for angles in different quadrants?
2. How do you apply trigonometric identities for angles beyond 180 degrees?
3. Why does \(\sin(270^\circ - \theta)\) result in \(-\cos(\theta)\)?
4. What are the sine and cosine values for all special angles like \(30^\circ\), \(45^\circ\), and \(60^\circ\)?
5. How do you determine the sign of trigonometric functions in different quadrants?

**Tip**: For angles in the third quadrant (like \(240^\circ\)), sine is negative while cosine is also negative. Always refer to the ASTC (All Students Take Calculus) rule to remember the signs in each quadrant.

my working out was this what was wrong? sin 240 = sin(270-30) = -sin(30) = 1/2

This page explains the common mistake in solving sin(240°) by incorrectly applying trigonometric identities. It covers how to use sin(270° - θ) = -cos(θ) and why sin(240°) should be -√3/2, not 1/2. Suitable for students in Grades 10-12 studying advanced trigonometry.

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