Trigonometry: Question 1

Worked solution

Step 1: Recall the range of $\sin x$

For any $x$, $\sin x$ takes every value in the interval $-1 \le \sin x \le 1$.

Step 2: Track what $-2\sin x$ does to that range

Multiplying by $-2$ reverses and stretches the range: as $\sin x$ increases from $-1$ to $1$, the term $-2\sin x$ decreases from $2$ to $-2$. So: $-2 \le -2\sin x \le 2$

Step 3: Add the vertical shift of $3$

$y = 3 - 2\sin x$

Adding $3$ throughout shifts every bound up by $3$: $3 - 2 \le y \le 3 + 2 \quad\Longrightarrow\quad 1 \le y \le 5$

The minimum value of $y$ is $1$, occurring when $\sin x = 1$ (at $x = 90^\circ$), because the minus sign in front of $2\sin x$ means the largest value of $\sin x$ produces the smallest value of $y$.

Why the other options are wrong

• B ($5$): this is the correct calculation for the maximum, not the minimum, of $y$ (it occurs when $\sin x = -1$).
• C ($-2$): this is just the extreme value of the term $-2\sin x$ on its own, without adding the vertical shift of $3$.
• D ($3$): this ignores the $-2\sin x$ term completely and reports only the vertical shift.

Final answer

• The minimum value of $y$ is $\boxed{1}$, option A.