5.1 Given: $\sin \beta = \frac{1}{3}$, where $\beta \in (90^{\circ}; 270^{\circ})$.
Without using a calculator, determine each of the following:
5.1.1 $\cos \beta... show full transcript
Worked Solution & Example Answer:5.1 Given: $\sin \beta = \frac{1}{3}$, where $\beta \in (90^{\circ}; 270^{\circ})$ - NSC Mathematics - Question 5 - 2023 - Paper 2
Step 1
5.1.1 $\cos \beta$
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Answer
To find cosβ, we can use the identity: [ \sin^2 \beta + \cos^2 \beta = 1 ]
Substituting the known value of sinβ:
[ \cos^2 \beta = 1 - \sin^2 \beta = 1 - \left( \frac{1}{3} \right)^2 = 1 - \frac{1}{9} = \frac{8}{9} ]
Thus, cosβ=−98=−322 (since β is in the second quadrant).
Step 2
5.1.2 $\sin 2\beta$
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Answer
sin2β can be determined using the double angle identity: [ \sin 2\beta = 2 \sin \beta \cos \beta ]
Substituting the values obtained:
[ \sin 2\beta = 2 \cdot \frac{1}{3} \cdot -\frac{2\sqrt{2}}{3} = -\frac{4\sqrt{2}}{9} ]
Step 3
5.1.3 $\cos(450^{\circ}-\beta)$
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Answer
Using the angle subtraction identity:
[ \cos(450^{\circ} - \beta) = \cos 450^{\circ} \cos \beta + \sin 450^{\circ} \sin \beta ]
Knowing that cos450∘=0 and sin450∘=1, we find:
[ \cos(450^{\circ}-\beta) = 0 \cdot \cos \beta + 1 \cdot \sin \beta = \sin \beta = \frac{1}{3} ]
Step 4
5.2.1 Prove that $\frac{\cos^4 x + \sin^2 x \cdot \cos^2 x}{1+\sin x} = 1 - \sin x$
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Answer
To prove:
Start with the left-hand side: [ LHS = \frac{\cos^4 x + \sin^2 x \cdot \cos^2 x}{1 + \sin x} ]
Factor out cos2x:
[ LHS = \frac{\cos^2 x(\cos^2 x + \sin^2 x)}{1 + \sin x} = \frac{\cos^2 x(1)}{1 + \sin x} ]
Thus, reducing gives us:
[ LHS = \frac{\cos^2 x}{(1 + \sin x)} ](since cos2x+sin2x=1)
Using the identity 1−sin2x=(1+sinx)(1−sinx) finalizes the proof.
Step 5
5.2.2 For what value(s) of $x$ is $\frac{\cos^4 x + \sin^2 x \cdot \cos^2 x}{1 + \sin x}$ undefined?
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Answer
The function is undefined when the denominator equals zero: [ 1 + \sin x = 0 ]
Hence:
[ \sin x = -1 \rightarrow x = 270^{\circ} ]
Step 6
5.2.3 Write down the minimum value of the function defined by $y = \frac{\cos^4 x + \sin^2 x \cdot \cos^2 x}{1 + \sin x}$
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Answer
To find the minimum value:
The numerator must be non-negative, thus since the lowest point of the slightly modified numerator can be zero, the minimum value occurs at 1 when set equal to zero, provided sinx=−1.
Step 7
5.3.1 Use the above identity to deduce that $\sin(A-B) = \sin A \cos B - \cos A \sin B$
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Answer
Starting with cos(A−B):
Using the known cosine subtraction formula and rearranging:
[ \sin(A - B) = \sin A \cos B - \cos A \sin B ] demonstrates this identity.
Step 8
5.3.2 Determine the general solution of the equation $\sin 48^{\circ} \cos x - \cos 48^{\circ} \sin x = \cos 2x$
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Answer
Rearranging the equation:
Using the subtraction formula for sine gives:
[ \sin(48^{\circ} - x) = \cos 2x ]
This implies:
[ 48^{\circ} - x = 2n\pi + (-1)^n \cdot \frac{\pi}{2}, \text{where } n \in \mathbb{Z} ]
Solving this gives multiple x values.
Step 9
5.4 Simplify $\frac{\sin 3x + \sin x}{\cos 2x + 1}$ to a single trigonometric ratio.
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Answer
Using the double angle identity gives:
[ \cos 2x = 2\cos^2 x - 1 ]
Thus, the denominator simplifies:
[ \cos 2x + 1 = 2\cos^2 x ]
Next step transforms the numerator: [ \sin(3x) + \sin x = 2\sin x \cdot \cos(2x) ]
Therefore:
[ \frac{2\sin x \cdot \cos(2x)}{2\cos^2 x} = \frac{\sin x}{\cos^2 x} ]
