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5.1 Given: sin 2x = \frac{\sqrt{15}}{8} and 0^{\circ} \leq 2x \leq 90^{\circ} - NSC Mathematics - Question 5 - 2017 - Paper 2
Question 5
5.1 Given: sin 2x = \frac{\sqrt{15}}{8} and 0^{\circ} \leq 2x \leq 90^{\circ}. Determine with the aid of a diagram and without using a calculator the value of cos ... show full transcript
Worked Solution & Example Answer:5.1 Given: sin 2x = \frac{\sqrt{15}}{8} and 0^{\circ} \leq 2x \leq 90^{\circ} - NSC Mathematics - Question 5 - 2017 - Paper 2
Step 1
5.1 Determine with the aid of a diagram and without using a calculator the value of cos x.
Answer
To find cos x, we start by drawing a right triangle. Given that sin 2x = \frac{\sqrt{15}}{8}, we will consider the opposite side length to be \sqrt{15}$ and the hypotenuse to be 8. Using the Pythagorean theorem, we can find the adjacent side length:
[ a^2 + b^2 = c^2 \implies a^2 + (\sqrt{15})^2 = 8^2 \implies a^2 + 15 = 64 \implies a^2 = 49 \implies a = 7. ]
Thus, we can use the adjacent side (7) and the hypotenuse (8) to find cos 2x:
[ cos 2x = \frac{adjacent}{hypotenuse} = \frac{7}{8} ]
To find cos x, we will use the identity ( cos 2x = 2 \cos^2 x - 1 ):
[ \frac{7}{8} = 2 \cos^2 x - 1 \implies 2 \cos^2 x = \frac{15}{8} \implies \cos^2 x = \frac{15}{16} \implies \cos x = \frac{\sqrt{15}}{4}. ]
Step 2
5.2 Simplify the following expression into one trigonometric ratio of \theta.
Answer
To simplify \frac{sin(180^{\circ} - \theta) \cdot sin(540^{\circ} - \theta) \cdot cos(\theta - 90^{\circ})}{tan(\theta) \cdot sin^{2}(360^{\circ} - \theta)}, we start by applying trigonometric identities:
- ( sin(180^{\circ} - \theta) = sin \theta )
- ( sin(540^{\circ} - \theta) = sin(180^{\circ} + 360^{\circ} - \theta) = -sin \theta )
- ( cos(\theta - 90^{\circ}) = -sin \theta )
- ( sin(360^{\circ} - \theta) = -sin \theta )
Now substituting these identities back into the equation gives: [ \frac{(sin \theta)(-sin \theta)(-sin \theta)}{tan(\theta)(-sin \theta)^{2}} = \frac{sin^{3} \theta}{tan(\theta)\cdot sin^{2} \theta} = \frac{sin \theta}{tan(\theta)} = cos \theta. ]
Step 3
5.3.1 Prove the above identity.
Answer
We need to prove the identity: \frac{sin 5x \cdot cos 3x - cos 5x \cdot sin 3x}{tan 2x} = 1 - 2 sin^{2} x.
Using the angle difference formula:
LHS = \frac{sin(5x - 3x)}{tan 2x} = \frac{sin(2x)}{tan(2x)} = \frac{sin(2x)}{\frac{sin(2x)}{cos(2x)}} = cos(2x).
Now, we can use the double angle identity: ( cos(2x) = 1 - 2 sin^{2} x ) so therefore, LHS = RHS.
Step 4
5.3.2 For which value(s) of x will the above identity be undefined for 0^{\circ} \leq x \leq 180^{\circ}?
Answer
The identity will be undefined when tan 2x equals 0, which occurs when 2x = 0^{\circ}, 180^{\circ}, or 360^{\circ}. Thus, dividing by 2 gives the solutions:
x = 0^{\circ}, 90^{\circ}.
Therefore, the identity is undefined for these values of x within the interval [0, 180].