$\frac12=-\frac12$

A sequence of equalities is given below, leading to the statement $\frac12=-\frac12$, which is obviously false. One of these equalities is incorrect. $$\frac12=\sin30^\circ\stackrel{(1)}=-\sin210^\circ\stackrel{(2)}=\sin150^\circ\stackrel{(3)}=-\cos120^\circ\stackrel{(4)}=\cos\left(-120^\circ\right)\stackrel{(5)}=-\cos60^\circ=-\frac12$$ The teacher asked the students to give a justification for the validity of each step. Here are their comments:

Alice: Equality (1) holds. For every real number $x$, $\sin ⁡x=-\sin\left(x+180^\circ\right)$. Therefore, the angles $30^\circ$ (from the 1st quadrant) and $210^\circ$ (from the 3rd quadrant) have opposite sine values.

Ben: Equality (2) holds. For every real number $x$, $\sin⁡\left(180^\circ+x\right)=-\sin\left(180^\circ-x\right)$. Therefore, the angles $210^\circ$ (from the 3rd quadrant) and $150^\circ$ (from the 2nd quadrant) have opposite sine function values.

Daniel: Equality (3) holds.

• For every real number $x$, $\sin x=\cos\left(x-90^\circ\right)$. Therefore, $\sin150^\circ=\cos⁡60^\circ$.
• For every real number $x$, $\cos ⁡x=-\cos\left(180^\circ-x\right)$. Therefore, the angles $60^\circ$ (from the 1st quadrant) and $120^\circ$ (from the 2nd quadrant) have opposite cosine values, i.e., $\cos 60^\circ=-\cos⁡120^\circ$.

Elena: Equality (4) holds. For every real number $x$, the following holds: $-\cos ⁡x=\cos⁡\left(-x\right)$. Therefore, $-\cos120^\circ=\cos\left(-120^0\right)$.

Filip: Equality (5) holds.

• The cosine function is periodic with a period of $360^\circ$. Therefore, $\cos\left(-120^\circ\right)=\cos⁡240^\circ$.
• For every real number $x$, $\cos x=-\cos\left(180^\circ+x\right)$. Therefore, the angles $60^\circ$ (from the 1st quadrant) and $240^\circ$ (from the 3rd quadrant) have opposite cosine values, i.e., $\cos 240^\circ=-\cos 60^\circ$.

Who is not right?