A-Level AQA Maths Pure Trigonometric Functions: 5 (a) Sketch the graph of $y =

5 (a) Sketch the graph of $y = ext{sin} \, 2x$ for $0^{ ext{°}} \leq x \leq 360^{ ext{°}}$ 5 (b) The equation 5 (b) The equation $\text{sin} \, 2x = A$ has exactly two solutions for $0^{ ext{°}} \leq x \leq 360^{ ext{°}}$ State the possible values of $A$. - AQA - A-Level Maths Pure - Question 5 - 2022 - Paper 3

Question 5

$5-(a)-Sketch-the-graph-of---$y-=--ext{sin}-\,-2x$---for-$0^{-ext{°}}-\leq-x-\leq-360^{-ext{°}}$---5-(b)-The-equation---5-(b)-The-equation---$\text{sin}-\,-2x-=-A$---has-exactly-two-solutions-for-$0^{-ext{°}}-\leq-x-\leq-360^{-ext{°}}$---State-the-possible-values-of-$A$.-AQA-A-Level Maths Pure-Question 5-2022-Paper 3.png$

5 (a) Sketch the graph of $y = ext{sin} \, 2x$ for $0^{ ext{°}} \leq x \leq 360^{ ext{°}}$ 5 (b) The equation 5 (b) The equation $\text{sin} \, 2x = A$ ... show full transcript

Worked Solution & Example Answer:5 (a) Sketch the graph of $y = ext{sin} \, 2x$ for $0^{ ext{°}} \leq x \leq 360^{ ext{°}}$ 5 (b) The equation 5 (b) The equation $\text{sin} \, 2x = A$ has exactly two solutions for $0^{ ext{°}} \leq x \leq 360^{ ext{°}}$ State the possible values of $A$. - AQA - A-Level Maths Pure - Question 5 - 2022 - Paper 3

Step 1

Sketch the graph of $y = \text{sin} \, 2x$

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Answer

To sketch the graph of the function $y = \text{sin} \, 2x$ , we need to understand its behavior over the interval from $0^{\text{°}}$ to $360^{\text{°}}$ . The sine function has a periodicity of $360^{\text{°}}$ , but because of the factor of 2 in the argument, the period in terms of $x$ is halved to $180^{\text{°}}$ .

Key Points:

Amplitude: The amplitude is 1, so the graph will oscillate between -1 and 1.
Zeros: The graph intersects the x-axis at multiples of $90^{\text{°}}$ : $0^{\text{°}}$ , $90^{\text{°}}$ , $180^{\text{°}}$ , $270^{\text{°}}$ , and $360^{\text{°}}$ .
Peaks and Troughs: The graph reaches its maximum at $90^{\text{°}}$ (1) and its minimum at $270^{\text{°}}$ (-1).

The sketch should accurately depict these features, ensuring the correct orientation through the origin and capturing at least one full cycle of the sine wave.

Step 2

State the possible values of A.

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Answer

For the equation $\text{sin} \, 2x = A$ to have exactly two solutions in the interval $0^{\text{°}} \leq x \leq 360^{\text{°}}$ , the value of $A$ must lie within the range of the sine function.

Possible Values:

The sine function can take values from -1 to 1. Thus, to have exactly two solutions, $A$ must be within these bounds but not equal to the extreme values. Therefore, the possible values of $A$ are:

$-1 < A < 1$

In simpler terms, the values must satisfy:

eq -1 \text{ and } A \neq 1$$

Sketch the graph of $y = \text{sin} \, 2x$

Key Points:

State the possible values of A.

Possible Values:

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