14. (a) Express $ \sqrt{3} \sin x - \cos x $ in the form $ k \sin(x - \alpha) ^\circ $, where $ k > 0 $ and $ 0 < \alpha < 360 $. (b) Hence, or otherwise, sketch the graph with equation $ y = \sqrt{3} \sin x - \cos x $, $ 0 \leq x \leq 360 $. Use the diagram provided in the answer booklet.

Scottish Highers Maths Wave Function: 14. (a) Express \( \sqrt{3} \sin

14. (a) Express $ \sqrt{3} \sin x - \cos x $ in the form $ k \sin(x - \alpha) ^\circ $, where $ k > 0 $ and $ 0 < \alpha < 360 $ - Scottish Highers Maths - Question 14 - 2017

Question 14

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14. (a) Express $ \sqrt{3} \sin x - \cos x $ in the form $ k \sin(x - \alpha) ^\circ $, where $ k > 0 $ and $ 0 < \alpha < 360 $. (b) Hence, or otherwise, s... show full transcript

Worked Solution & Example Answer:14. (a) Express $ \sqrt{3} \sin x - \cos x $ in the form $ k \sin(x - \alpha) ^\circ $, where $ k > 0 $ and $ 0 < \alpha < 360 $ - Scottish Highers Maths - Question 14 - 2017

Step 1

Express $ \sqrt{3} \sin x - \cos x $ in the form $ k \sin(x - \alpha) $

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Answer

To express ( \sqrt{3} \sin x - \cos x ) in the required form, we can use the compound angle formula for sine:

[ \sin(A - B) = \sin A \cos B - \cos A \sin B ]

Here, we can compare coefficients:

Setting ( \sin A = \sqrt{3} ) and ( \cos A = -1 ), we see that ( k ) must equal ( 2 ).
This gives us:
- ( \sin \alpha = \frac{\sqrt{3}}{2} ) and ( \cos \alpha = -\frac{1}{2} \
The angle ( \alpha ) can then be determined as ( 210^\circ ), since both sine and cosine are derived from standard values.

Therefore, we can conclude: [ k = 2, \quad \alpha = 210^\circ ] Thus, the expression is: [ \sqrt{3} \sin x - \cos x = 2 \sin(x - 210^\circ) ]

Step 2

Sketch the graph with equation $ y = \sqrt{3} \sin x - \cos x $

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Answer

To sketch the graph, we begin by noting the amplitude and periodic behavior of the function:

The amplitude is equal to ( |k| = 2 ).
The period of the sine function is normally ( 360^\circ ), so it remains the same.
The function intersects the x-axis at points where ( y = 0 ). Using the transformed angle ( 210^\circ ), we find the critical points around this angle:
- The turning points occur primarily around ( 30^\circ ) and ( 210^\circ ).
The graph has roots at points identified from graphing the original equation, specifically at ( 30 ) and ( 210 ).
The coordinates of intersections can be marked on the graph at those calculated values.

This results in a sine wave that oscillates with a maximum of ( 2 ) and minimum of ( -2 ) crossing the x-axis at the calculated roots. Marking these points will help visualize the required graph.

14. (a) Express \( \sqrt{3} \sin x - \cos x \) in the form \( k \sin(x - \alpha) ^\circ \), where \( k > 0 \) and \( 0 < \alpha < 360 \) - Scottish Highers Maths - Question 14 - 2017

Worked Solution & Example Answer:14. (a) Express \( \sqrt{3} \sin x - \cos x \) in the form \( k \sin(x - \alpha) ^\circ \), where \( k > 0 \) and \( 0 < \alpha < 360 \) - Scottish Highers Maths - Question 14 - 2017

Express \( \sqrt{3} \sin x - \cos x \) in the form \( k \sin(x - \alpha) \)

Sketch the graph with equation \( y = \sqrt{3} \sin x - \cos x \)

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