Question

find an angle $\theta$ with $0^{circ}<\theta<360^{circ}$ that has the same: sine as $200^{circ}$: $\theta = ___$ degrees cosine as $200^{circ}$: $\theta = ___$ degrees

Explanation:

Step1: Recall sine - angle relationship

The sine function has the property $\sin\theta=\sin(180^{\circ}-\alpha)$ for $\theta$ in the range $0^{\circ}<\theta < 360^{\circ}$. Given $\alpha = 200^{\circ}$, we know that $\sin(200^{\circ})=\sin(180^{\circ}+ 20^{\circ})=-\sin(20^{\circ})$. Also, $\sin(160^{\circ})=\sin(180^{\circ}-20^{\circ})=\sin(20^{\circ})=-\sin(200^{\circ})$. Since the sine function has a period of $360^{\circ}$, and $\sin\theta=\sin(180^{\circ}-\theta)$ for $\theta\in[0^{\circ},180^{\circ}]$, for $\theta = 200^{\circ}$, the angle with the same sine value in the range $0^{\circ}<\theta<360^{\circ}$ is $160^{\circ}$.

Step2: Recall cosine - angle relationship

The cosine function has the property $\cos\theta=\cos(360^{\circ}-\alpha)$ for $\theta$ in the range $0^{\circ}<\theta < 360^{\circ}$. Given $\alpha = 200^{\circ}$, we know that $\cos(200^{\circ})=\cos(180^{\circ}+20^{\circ})=-\cos(20^{\circ})$. And $\cos(160^{\circ})=\cos(180^{\circ}- 20^{\circ})=-\cos(20^{\circ})$. Also, $\cos(160^{\circ})=\cos(360^{\circ}-200^{\circ})$. The angle with the same cosine value as $200^{\circ}$ in the range $0^{\circ}<\theta<360^{\circ}$ is $160^{\circ}$.

Answer:

Sine as $200^{\circ}$: $160$
Cosine as $200^{\circ}$: $160$