$\sin{(180^\circ-\theta)}$ identity in Trigonometric method

Express the angle in the form a standard angle

$\implies$ $\sin{(180^\circ-\theta)}$ $\,=\,$ $\sin{(90^\circ+90^\circ-\theta)}$

$\,\,\,\,\,\,=\,\,$ $\sin{\big(90^\circ+(90^\circ-\theta)\big)}$

Let us denote the ninety degrees minus theta by an angle alpha for our convenience. It means, $\alpha$ $\,=\,$ $90^\circ-\theta$. Now, replace the $90$ degrees minus theta by alpha in the trigonometric expression.

$\,\,\,\,\,\,=\,\,$ $\sin{\big(90^\circ+(\alpha)\big)}$

The sine of angle $180^\circ$ minus theta is now converted as the sine of angle $90^\circ$ plus alpha.

$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(180^\circ-\theta)}$ $\,=\,$ $\sin{(90^\circ+\alpha)}$

Therefore, the sine of one hundred eighty degrees minus theta can be evaluated by finding the value of the sine of angle ninety degrees plus alpha.

Evaluate sine of sum of angles by the expansion

The ninety degrees plus alpha is a compound angle and the sine of this compound angle represents the sine of sum of the angles. So, it can be expanded as per the angle sum trigonometric identity of sine function.

$\implies$ $\sin{(90^\circ+\alpha)}$ $\,=\,$ $\sin{(90^\circ)}\cos{\alpha}$ $+$ $\cos{(90^\circ)}\sin{\alpha}$

$\,\,\,\,\,\,=\,\,$ $\sin{(90^\circ)} \times \cos{\alpha}$ $+$ $\cos{(90^\circ)} \times \sin{\alpha}$

We know that the sin of ninety degrees is equal to one and the cos of ninety degrees is zero.

$\,\,\,\,\,\,=\,\,$ $1 \times \cos{\alpha}$ $+$ $0 \times \sin{\alpha}$

Let’s focus on simplifying the trigonometric expression to find the sine of angle ninety degrees plus alpha function.

$\,\,\,\,\,\,=\,\,$ $\cos{\alpha}$ $+$ $0$

$\,\,\,\,\,\,=\,\,$ $\cos{\alpha}$

Therefore, it is evaluated that the sine of angle ninety degrees plus alpha is equal to cosine of angle alpha.

$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(90^\circ+\alpha)}$ $\,=\,$ $\cos{\alpha}$

Evaluate sine of difference of angles by the expansion

$\sin{(180^\circ-\theta)}$ $\,=\,$ $\sin{(90^\circ+\alpha)}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(180^\circ-\theta)}$ $\,=\,$ $\cos{\alpha}$

$\,\,\,\therefore\,\,\,\,\,\,$ $\sin{(180^\circ-\theta)}$ $\,=\,$ $\cos{(90^\circ-\theta)}$

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