In the diagram below, the graphs of $f(x) = ext{cos} 2x$ and $g(x) = - ext{sin} x$ are drawn for the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$. A and B are two points of intersection of $f$ and $g$. 6.1 Without using a calculator, determine the values of $x$ for which $ ext{cos} 2x = - ext{sin} x$ in the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$. 6.2 Use the graphs above to answer the following questions: 6.2.1 How many degrees apart are points A and B from each other? 6.2.2 For which values of $x$ in the given interval will $f^{ ext{'}}(x) > 0$? 6.2.3 Determine the values of $k$ for which $ ext{cos} 2x + 3 = k$ will have no solution.

NSC Mathematics Trigonometry: In the diagram below, the graphs

In the diagram below, the graphs of $f(x) = ext{cos} 2x$ and $g(x) = - ext{sin} x$ are drawn for the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2021 - Paper 2

Question 6

$In-the-diagram-below,-the-graphs-of-$f(x)-=--ext{cos}-2x$-and-$g(x)-=---ext{sin}-x$-are-drawn-for-the-interval-$x--orall-[-180^{-ext{o}}-;-180^{-ext{o}}]$-NSC Mathematics-Question 6-2021-Paper 2.png$

In the diagram below, the graphs of $f(x) = ext{cos} 2x$ and $g(x) = - ext{sin} x$ are drawn for the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$. A and B a... show full transcript

Worked Solution & Example Answer:In the diagram below, the graphs of $f(x) = ext{cos} 2x$ and $g(x) = - ext{sin} x$ are drawn for the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2021 - Paper 2

Step 1

Without using a calculator, determine the values of $x$ for which $ ext{cos} 2x = - ext{sin} x$ in the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$.

96%

114 rated

Answer

To solve the equation $ext{cos} 2x + ext{sin} x = 0$ , we can rewrite it using the identity:

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

This leads to the quadratic equation:

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

Using the quadratic formula:

$ext{sin} x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Substituting $a = 2$ , $b = 1$ , and $c = -1$ gives us:

$ext{sin} x = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4}$

Thus, $ext{sin} x = \frac{1}{2}$ or $ext{sin} x = -1$ . Values for $ext{sin} x = \frac{1}{2}$ in the given interval are:

$x = 30^{ ext{o}} + 360^{ ext{o}}k$ $x = 150^{ ext{o}} + 360^{ ext{o}}k$

For $ext{sin} x = -1$ :

$x = -90^{ ext{o}} + 360^{ ext{o}}k$

Thus, the values of $x$ are $x = 150^{ ext{o}}, 30^{ ext{o}}, -90^{ ext{o}}$ and similar solutions derived from periodicity in the given domain.

Step 2

6.2.1 How many degrees apart are points A and B from each other?

99%

104 rated

Answer

To determine the distance between points A and B, we note the angles where the functions intersect. Using the previously calculated angle values, we find:

$AB = 150^{ ext{o}} - (-30^{ ext{o}}) = 180^{ ext{o}}$

Therefore, points A and B are 180 degrees apart.

Step 3

6.2.2 For which values of $x$ in the given interval will $f^{ ext{'}}(x) > 0$?

96%

101 rated

Answer

The derivative of $f(x) = ext{cos} 2x$ is:

$f^{ ext{'}}(x) = -2 ext{sin} 2x$

To find when this is greater than zero, we look for values of $x$ such that:

$-2 ext{sin} 2x > 0 \Rightarrow ext{sin} 2x < 0$

The intervals where $ext{sin} 2x < 0$ occur in Quadrants 3 and 4. Therefore, we can identify the appropriate ranges within the domain to find:

$x ext{ in } (90^{ ext{o}}, 180^{ ext{o}}) ext{ and } (-180^{ ext{o}}, -90^{ ext{o}})$

Step 4

6.2.3 Determine the values of $k$ for which $ ext{cos} 2x + 3 = k$ will have no solution.

98%

120 rated

Answer

Considering the range of $ext{cos} 2x$ , which oscillates between -1 and 1, we modify the equation:

$ext{cos} 2x + 3 = k \Rightarrow k = ext{cos} 2x + 3$

Thus, the values of $k$ for which there will be no solution occurs when:

$k < 2 ext{ or } k > 4$

This means for any $k$ less than 2 or greater than 4, $ext{cos} 2x + 3 = k$ will yield no valid solutions.

In the diagram below, the graphs of $f(x) = ext{cos} 2x$ and $g(x) = - ext{sin} x$ are drawn for the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2021 - Paper 2

Worked Solution & Example Answer:In the diagram below, the graphs of $f(x) = ext{cos} 2x$ and $g(x) = - ext{sin} x$ are drawn for the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$ - NSC Mathematics - Question 6 - 2021 - Paper 2

Without using a calculator, determine the values of $x$ for which $ ext{cos} 2x = - ext{sin} x$ in the interval $x orall [-180^{ ext{o}} ; 180^{ ext{o}}]$.

6.2.1 How many degrees apart are points A and B from each other?

6.2.2 For which values of $x$ in the given interval will $f^{ ext{'}}(x) > 0$?

6.2.3 Determine the values of $k$ for which $ ext{cos} 2x + 3 = k$ will have no solution.

Join the NSC students using SimpleStudy...