In the diagram, the graph of $f(x) = ext{cos} 2x$ is drawn for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$. 6.1 Draw the graph of $g(x) = 2 ext{sin} x - 1$ for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$ on the grid given in your ANSWER BOOK. Show ALL the intercepts with the axes, as well as the turning points. 6.2 Let A be a point of intersection of the graphs of $f$ and $g$. Show that the $x$-coordinate of A satisfies the equation $ ext{sin} x = rac{-1+ ext{√}5}{2}$. 6.3 Hence, calculate the coordinates of the points of intersection of graphs of $f$ and $g$ for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$.

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In the diagram, the graph of $f(x) = ext{cos} 2x$ is drawn for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2017 - Paper 2

Question 6

$In-the-diagram,-the-graph-of--$f(x)-=--ext{cos}-2x$-is-drawn-for-the-interval-$x--ext{-in-}-[-270^{-ext{o}};90^{-ext{o}}]$-NSC Mathematics-Question 6-2017-Paper 2.png$

In the diagram, the graph of $f(x) = ext{cos} 2x$ is drawn for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$. 6.1 Draw the graph of $g(x) = 2 ext{sin}... show full transcript

Worked Solution & Example Answer:In the diagram, the graph of $f(x) = ext{cos} 2x$ is drawn for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2017 - Paper 2

Step 1

Draw the graph of $g(x) = 2 ext{sin} x - 1$

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Answer

To draw the graph of $g(x) = 2 ext{sin} x - 1$ , we start by identifying key points and intercepts:

Amplitude and Period: The amplitude is 2, and the period is $360^{ ext{o}}$ . The function completes its cycle between $[-270^{ ext{o}};90^{ ext{o}}]$ .
Key Points: Calculate the y-values for critical x-values:
- At $x = -270^{ ext{o}}$ , ( g(-270^{ ext{o}}) = 2 ext{sin}(-270^{ ext{o}}) - 1 = 2(-1) - 1 = -3 )
- At $x = -180^{ ext{o}}$ , ( g(-180^{ ext{o}}) = 2 ext{sin}(-180^{ ext{o}}) - 1 = -1 )
- At $x = -90^{ ext{o}}$ , ( g(-90^{ ext{o}}) = 2 imes (-1) - 1 = -3 )
- At $x = 0$ , ( g(0) = 2 ext{sin}(0) - 1 = -1 )
- At $x = 90^{ ext{o}}$ , ( g(90^{ ext{o}}) = 2(1) - 1 = 1 )
Plotting the Graph: Plot the points: (-270, -3), (-180, -1), (0, -1), (90, 1) and sketch a smooth curve to represent the sinusoidal behavior.
Turning Points: Identify the maxima and minima. For the given interval, the turning point occurs at $x = -90^{ ext{o}}$ and $x = 90^{ ext{o}}$ .

Step 2

Let A be a point of intersection of the graphs of $f$ and $g$. Show that the $x$-coordinate of A satisfies the equation $ ext{sin} x = rac{-1+ ext{√}5}{2}$

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Answer

To find the points of intersection between the graphs of $f$ and $g$ , we set:

$ext{cos} 2x = 2 ext{sin} x - 1$

Using the identity for $ext{cos} 2x$ :

$1 - 2 ext{sin}^2 x = 2 ext{sin} x - 1$

Rearranging gives:

$2 ext{sin}^2 x + 2 ext{sin} x - 2 = 0$

Dividing through by 2:

$ext{sin}^2 x + ext{sin} x - 1 = 0$

This is a quadratic equation in terms of $ext{sin} x$ . We can use the quadratic formula:

$ext{sin} x = rac{-b ext{±} ext{sqrt}(b^2 - 4ac)}{2a}$

where $a = 1$ , $b = 1$ , $c = -1$ . Thus:

$ext{sin} x = rac{-1 ext{±} ext{sqrt}(1^2 - 4 imes 1 imes (-1))}{2 imes 1}$

Simplifying:

$ext{sin} x = rac{-1 ext{\pm} ext{\sqrt5}}{2}$

The solution corresponding to $A$ is:

$ext{sin} x = rac{-1 + ext{\sqrt}5}{2}$

Step 3

Hence, calculate the coordinates of the points of intersection of graphs of $f$ and $g$ for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$

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Answer

To find the coordinates of the intersection points using the solution found in the previous step:

From the equation ( ext{sin} x = rac{-1 + ext{√}5}{2} ):
- Approximate the value: ( rac{-1 + ext{√}5}{2} ext{ is approximately } 0.618. )
Finding x-values: The general solutions where ( ext{sin} x = 0.618 ) are:
- ( x = ext{sin}^{-1}(0.618) ext{ and } 180^{ ext{o}} - ext{sin}^{-1}(0.618) )
Calculate the angles:
- ( x_1 ext{ is approximately } 38.17^{ ext{o}} )
- ( x_2 = 141.83^{ ext{o}} )
- Additionally, we find the corresponding values for negative angles: ( x_3 ext{ is approximately } 218.17^{ ext{o}} )
Coordinates: For each x-value, use $f(x)$ or $g(x)$ to find y-coordinates:
- For ( 38.17^{ ext{o}} ): ( f(38.17^{ ext{o}}) ext{ gives } (38.17, 0.24) )
- For ( 218.17^{ ext{o}} ): ( f(218.17^{ ext{o}}) ext{ gives } (218.17, 0.24) )
Final Coordinates: The points of intersection are approximately:
- $(38.17, 0.24)$ and $(218.17, 0.24)$ .

In the diagram, the graph of $f(x) = ext{cos} 2x$ is drawn for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2017 - Paper 2

Worked Solution & Example Answer:In the diagram, the graph of $f(x) = ext{cos} 2x$ is drawn for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2017 - Paper 2

Draw the graph of $g(x) = 2 ext{sin} x - 1$

Let A be a point of intersection of the graphs of $f$ and $g$. Show that the $x$-coordinate of A satisfies the equation $ ext{sin} x = rac{-1+ ext{√}5}{2}$

Hence, calculate the coordinates of the points of intersection of graphs of $f$ and $g$ for the interval $x ext{ in } [-270^{ ext{o}};90^{ ext{o}}]$

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