The sine of sum of the angles ninety degrees and theta is equal to cosine of angle theta.

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

The sine of ninety degrees plus theta formula can be derived mathematically by a trigonometric identity and the values of sine and cosine functions.

Now, let’s learn how to prove the sine of the sum of angles ninety degrees plus theta identity in mathematical form by trigonometry.

The trigonometric expression $\sin{(90^\circ+\theta)}$ expresses the sine of sum of two angles. So, it can be expanded by the angle sum identity of sine function.

$\sin{(A+B)}$ $\,=\,$ $\sin{A}\cos{B}$ $+$ $\cos{A}\sin{B}$

Assume $A \,=\, 90^\circ$ and $B \,=\, \theta$. Now, substitute them in the expansion of the angle sum identity of sine function.

$=\,\,\,$ $\sin{90^\circ}\cos{\theta}$ $+$ $\cos{90^\circ}\sin{\theta}$

$=\,\,\,$ $\sin{90^\circ} \times \cos{\theta}$ $+$ $\cos{90^\circ} \times \sin{\theta}$

According to the trigonometry, the sin of 90 degrees is one and cos of 90 degrees is zero. So, substitute them in the trigonometric expression to find its value.

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

It is time to evaluate the trigonometric expression and it can be calculated by simplification.

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

$=\,\,\,$ $\cos{\theta}$

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

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

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