Write the equations of the tangent and normal to each

Write the equations of the tangent and normal to each curve at the points $x=0$ and $x=\dfrac{\pi}{2}$: 1° $y=\sin x$; 2° $y=\cos x$.

Step 1: $1^{\circ}$: $y=\sin x$

$y'=\cos x$.

At $x=0$: $y(0)=0$, $y'(0)=1$; point $(0,0)$, tangent slope $1$, normal slope $-1$.

\[ \text{Tangent: } y=x; \qquad \text{Normal: } y=-x. \tag{1} \]

At $x=\dfrac\pi2$: $y\!\left(\tfrac\pi2\right)=1$, $y'\!\left(\tfrac\pi2\right)=0$; the tangent is horizontal and the normal vertical.

\[ \text{Tangent: } y=1; \qquad \text{Normal: } x=\frac\pi2. \tag{2} \]

Step 2: $2^{\circ}$: $y=\cos x$

$y'=-\sin x$.

At $x=0$: $y(0)=1$, $y'(0)=0$; tangent horizontal, normal vertical.

\[ \text{Tangent: } y=1; \qquad \text{Normal: } x=0. \tag{3} \]

At $x=\dfrac\pi2$: $y\!\left(\tfrac\pi2\right)=0$, $y'\!\left(\tfrac\pi2\right)=-1$; tangent slope $-1$, normal slope $1$.

\[ \text{Tangent: } y=-x+\frac\pi2; \qquad \text{Normal: } y=x-\frac\pi2. \tag{4} \]

Step 3: Verification

\[\text{Each tangent/normal pair is perpendicular: product of the two slopes } = -1 \text{ wherever both are finite}\ \checkmark\]

\[\text{Each line passes through its stated point: direct substitution in (1)–(4)}\ \checkmark\]