Question

in the triangle, which is equivalent to sin a?

Explanation:

Step1: Recall trigonometric relations in right - triangle

In a right - triangle, $\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}$ and $\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}$. Let the right - triangle have angles $A$, $B$, and $C$ with $C = 90^{\circ}$. Then $A + B=90^{\circ}$, so $B = 90^{\circ}-A$.

Step2: Analyze the relationship between sine and cosine

We know that $\sin A=\frac{\text{opposite side to }A}{\text{hypotenuse}}$ and $\cos C=\cos90^{\circ}=0$, $\sin B=\sin(90^{\circ}-A)=\cos A$, $\sin C=\sin90^{\circ} = 1$. Also, $\sin A=\cos(90^{\circ}-A)$. In a right - triangle, if $C = 90^{\circ}$, then $\sin A=\cos B$. And since $B=90^{\circ}-C$, we have $\sin A=\cos C$.

Answer:

$\cos C$