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INSTRUCTIONS

SOROTI SECONDARY SCHOOL

END OF TERM ONE EXAMINATION 2024

S.6 MATHEMATICS PAPER 1

TIME: 3 HOURS

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Answer all questions in Section A and any FIVE questions from Section B

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SECTION A

1. The [tex]$5^{\text{th}}$[/tex] term of an Arithmetic progression (A.P) is 1.2 and the sum of the first 5 terms is 80. Determine the first term and the common difference.
(5 marks)

2. Given that [tex]$\sin \left(\theta - 45^{\circ}\right) = 3 \cos \left(\theta + 45^{\circ}\right)$[/tex], show that [tex]$\tan \theta = 1$[/tex]. Hence, find [tex]$\theta$[/tex] if [tex]$0^{\circ} \leq \theta \leq 360^{\circ}$[/tex].

Sagot :

Sure, let's solve the problem step-by-step.

Given Equation: [tex]\( \sin(\theta - 45^\circ) = 3 \cos(\theta + 45^\circ) \)[/tex]

1. Apply the sine and cosine angle subtraction and addition formulas:
[tex]\[ \sin(a - b) = \sin(a) \cos(b) - \cos(a) \sin(b) \][/tex]
[tex]\[ \cos(a + b) = \cos(a) \cos(b) - \sin(a) \sin(b) \][/tex]

2. Rewrite the given equation using these formulas:
[tex]\[ \sin(\theta - 45^\circ) = \sin(\theta) \cos(45^\circ) - \cos(\theta) \sin(45^\circ) \][/tex]
[tex]\[ \cos(\theta + 45^\circ) = \cos(\theta) \cos(45^\circ) - \sin(\theta) \sin(45^\circ) \][/tex]

3. Note that [tex]\( \cos(45^\circ) = \sin(45^\circ) = \frac{\sqrt{2}}{2} \)[/tex]:
[tex]\[ \sin(\theta - 45^\circ) = \sin(\theta) \cdot \frac{\sqrt{2}}{2} - \cos(\theta) \cdot \frac{\sqrt{2}}{2} \][/tex]
[tex]\[ \cos(\theta + 45^\circ) = \cos(\theta) \cdot \frac{\sqrt{2}}{2} - \sin(\theta) \cdot \frac{\sqrt{2}}{2} \][/tex]

4. Substitute these expressions back into the given equation:
[tex]\[ \frac{\sqrt{2}}{2} \left( \sin(\theta) - \cos(\theta) \right) = 3 \cdot \frac{\sqrt{2}}{2} \left( \cos(\theta) - \sin(\theta) \right) \][/tex]

5. Simplify by multiplying through by [tex]\( 2/\sqrt{2} \)[/tex]:
[tex]\[ \sin(\theta) - \cos(\theta) = 3 \left( \cos(\theta) - \sin(\theta) \right) \][/tex]

6. Distribute the terms on the right-hand side:
[tex]\[ \sin(\theta) - \cos(\theta) = 3 \cos(\theta) - 3 \sin(\theta) \][/tex]

7. Combine like terms:
[tex]\[ \sin(\theta) + 3 \sin(\theta) = 3 \cos(\theta) + \cos(\theta) \][/tex]
[tex]\[ 4 \sin(\theta) = 4 \cos(\theta) \][/tex]

8. Divide both sides by 4:
[tex]\[ \sin(\theta) = \cos(\theta) \][/tex]

9. Therefore, [tex]\( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = 1 \)[/tex]:

10. Solve for [tex]\( \theta \)[/tex]:

The angles where [tex]\( \tan(\theta) = 1 \)[/tex] within [tex]\( 0^\circ \leq \theta \leq 360^\circ \)[/tex] are:
[tex]\[ \theta = 45^\circ + k \cdot 180^\circ, \quad \text{where } k \text{ is any integer} \][/tex]

For the given range [tex]\( 0^\circ \leq \theta \leq 360^\circ \)[/tex]:
- When [tex]\( k = 0 \)[/tex]: [tex]\( \theta = 45^\circ \)[/tex]
- When [tex]\( k = 1 \)[/tex]: [tex]\( \theta = 225^\circ \)[/tex]

No other values of [tex]\( k \)[/tex] within the given range fit. Therefore, the solution is:

[tex]\[ \theta = 45^\circ \text{ and } 225^\circ \][/tex]

Thus, the values of [tex]\( \theta \)[/tex] that satisfy the given equation are [tex]\( 45^\circ \)[/tex] and [tex]\( 225^\circ \)[/tex].