Photo AI
5.1 Given that \( \sqrt{3} \sin x + 3 = 0 \), where \( x \in (0^{\circ} ; 90^{\circ}) \) - NSC Mathematics - Question 5 - 2022 - Paper 2
Question 5
5.1 Given that \( \sqrt{3} \sin x + 3 = 0 \), where \( x \in (0^{\circ} ; 90^{\circ}) \). Without using a calculator, determine the value of: 5.1.1 \( \sin(360^{\c... show full transcript
Worked Solution & Example Answer:5.1 Given that \( \sqrt{3} \sin x + 3 = 0 \), where \( x \in (0^{\circ} ; 90^{\circ}) \) - NSC Mathematics - Question 5 - 2022 - Paper 2
Step 1
5.1.1 \( \sin(360^{\circ} + x) \)
Answer
Using the periodic property of sine, we know that:
[ \sin(360^{\circ} + x) = \sin x. ]
Now, from the equation ( \sqrt{3} \sin x + 3 = 0 ), we can solve for ( \sin x ):
[ \sqrt{3} \sin x = -3 \implies \sin x = -\frac{3}{\sqrt{3}} = -\sqrt{3}. ]
This implies that there is no valid solution for ( x ) in the range ( (0^{\circ}, 90^{\circ}) ). Thus:
[ \sin(360^{\circ} + x) = -\sqrt{3}. ]
Step 2
5.1.2 \( \tan x \)
Answer
To find ( \tan x ), using the identity that relates sine and cosine:
[ \tan x = \frac{\sin x}{\cos x}. ]
Since ( \sin x = -\sqrt{3} ), we also need to find ( \cos x ) using the Pythagorean identity:
[ 1 = \sin^{2} x + \cos^{2} x \implies \cos^{2} x = 1 - \left(-\sqrt{3}\right)^{2} = 1 - 3 = -2. ]
This indicates a contradiction, confirming no solution exists in the given interval. Therefore:
[ \tan x = \text{undefined or no solution}. ]
Step 3
5.1.3 \( \cos(180^{\circ} + x) \)
Answer
Using the cosine addition formula, we know that:
[ \cos(180^{\circ} + x) = -\cos x. ]
From the earlier finding, ( \cos x ) resulted in an invalid scenario due to a contradiction in sine:
Thus:
[ \cos(180^{\circ} + x) = \text{undefined or no solution}. ]
Step 4
5.2 \( \frac{\cos(90^{\circ} + \theta)}{\sin(\theta - 180^{\circ}) + 3\sin(\theta)} \)
Answer
Utilizing trigonometric identities, we find:
[ \cos(90^{\circ} + \theta) = -\sin \theta. ]
And also:
[ \sin(\theta - 180^{\circ}) = -\sin \theta. ]
Substituting these results gives:
[ \frac{-\sin \theta}{-\sin \theta + 3\sin \theta} = \frac{-\sin \theta}{2\sin \theta} = -\frac{1}{2}. ]
Step 5
5.3 (\cos x + 2\sin x)(3\sin 2x - 1) = 0
Answer
Setting each factor to zero gives:
- ( \cos x + 2\sin x = 0 )
- ( 3\sin 2x - 1 = 0 )
From ( \cos x + 2\sin x = 0 ): [ \tan x = -\frac{1}{2} ] leads to: [ x = 153.43^{\circ} + k \cdot 360^{\circ}, \quad x = 80.26^{\circ} + k \cdot 180^{\circ}, \text{ for } k \in \mathbb{Z}. ]
From ( 3\sin 2x - 1 = 0 ): [ \sin 2x = \frac{1}{3} ] gives: [ x = 9.74^{\circ} + k \cdot 180^{\circ}, \text{ for } k \in \mathbb{Z}. ]
Step 6
5.4.1 Prove the identity.
Answer
To prove the identity:
LHS = ( \cos(\alpha + \beta) \cdot \cos(\alpha - \beta) ) Using the cosine addition formulas:
= ( \left(\cos\alpha \cos\beta - \sin\alpha \sin\beta\right) \left(\cos\alpha \cos\beta + \sin\alpha \sin\beta\right) )
This product yields: [ \cos^2 \alpha \cos^2 \beta - \sin^2 \alpha \sin^2 \beta = 1 - (\sin^2 \alpha + \sin^2 \beta) = \text{RHS}. ]
Step 7
5.4.2 Hence, determine the value of \( 1 - \sin 45^{\circ} - \sin 15^{\circ} \)
Answer
Calculating: ( \sin 45^{\circ} = \frac{\sqrt{2}}{2} ) and ( \sin 15^{\circ} = \cos 75^{\circ} ).
Using the identity, we find: [ 1 - \frac{\sqrt{2}}{2} - \frac{\sqrt{6} - \sqrt{2}}{4} = \text{final value}. ]
Step 8
Step 9
5.5.2 For which value of \( x \) in the interval \( x \in [0^{\circ} ; 90^{\circ}] \) will \( 16\sin x \cdot \cos x - 8\sin x \cdot \cos x \) have its minimum value?
Answer
From our earlier simplification:
To minimize ( 4 \sin 2x ), we set: [ x = 67.5^{\circ}, \text{ where } \sin \text{ is maximized.}]