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6.1 Without using a calculator, simplify the following expression to a single trigonometric term - NSC Mathematics - Question 6 - 2022 - Paper 2

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6.1 Without using a calculator, simplify the following expression to a single trigonometric term. $$\frac{\sin 10^\circ}{\cos 440^\circ + \tan(360^\circ - \theta) \... show full transcript

Worked Solution & Example Answer:6.1 Without using a calculator, simplify the following expression to a single trigonometric term - NSC Mathematics - Question 6 - 2022 - Paper 2

Step 1

Without using a calculator, simplify the following expression to a single trigonometric term.

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Answer

To simplify the expression, start with:

sin10cos440+tan(360θ)sin20\frac{\sin 10^\circ}{\cos 440^\circ + \tan(360^\circ - \theta) \cdot \sin 20^\circ}

Step 1: Reduce angles
Firstly, simplify the angles in the expression. We know that:

  • (\cos 440^\circ = \cos(440^\circ - 360^\circ) = \cos 80^\circ).
  • (\tan(360^\circ - \theta) = -\tan \theta).

Using these results, the expression becomes:

sin10cos80tanθsin20\frac{\sin 10^\circ}{\cos 80^\circ - \tan \theta \cdot \sin 20^\circ}

Step 2: Utilize trigonometric identities
Now, recall the identity (\tan \theta = \frac{\sin \theta}{\cos \theta}). Substituting gives:

sin10cos80sinθcosθsin20\frac{\sin 10^\circ}{\cos 80^\circ - \frac{\sin \theta}{\cos \theta} \cdot \sin 20^\circ}

Step 3: Common denominator
Multiply through by (\cos \theta):

sin10cosθcos80cosθsinθsin20\frac{\sin 10^\circ \cos \theta}{\cos 80^\circ \cos \theta - \sin \theta \sin 20^\circ}

Using the angle sum and difference identity yields:

sin(20)=sin(10+10)=2sin10cos10.\sin(20^\circ) = \sin(10^\circ + 10^\circ) = 2 \sin 10^\circ \cos 10^\circ.

Therefore, this results in:

cos(20)=12sin2(10).\cos(20^\circ) = 1 - 2\sin^2(10^\circ).

Step 4: Final expression
After simplification, we have:

cos20\boxed{\cos 20^\circ}.

Step 2

Calculate the value of k if sin(60° + 2x) + sin(60° - 2x) = k cos 2x.

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Answer

To find the value of (k):

Step 1: Apply the sine addition formulas
Using the sine angle addition formula, we can rewrite the left-hand side:

sin(60+2x)+sin(602x)=2sin(60)cos(2x)\sin(60^\circ + 2x) + \sin(60^\circ - 2x) = 2 \sin(60^\circ) \cos(2x)

Step 2: Set equivalent expression
This leads to:

2sin(60)cos(2x)=kcos(2x)2 \sin(60^\circ) \cos(2x) = k \cos(2x)

Step 3: Solve for k
Dividing each side by (\cos(2x)) (assuming (\cos(2x) \neq 0)):

k=2sin(60) k = 2 \sin(60^\circ)

Using (\sin(60^\circ) = \frac{\sqrt{3}}{2}):

k=232=3. k = 2 \cdot \frac{\sqrt{3}}{2} = \sqrt{3}.

Thus, the final result is:

k=3\boxed{k = \sqrt{3}}.

Step 3

If cos x = sqrt{t}, without using a calculator, determine the value of tan 60°[sin(60° + 2x) + sin(60° - 2x)] in terms of t.

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Answer

To find the value:

Step 1: Rewrite the expression
We begin with:

tan60[sin(60+2x)+sin(602x)]=3[sin(60+2x)+sin(602x)]\tan 60^\circ [\sin(60^\circ + 2x) + \sin(60^\circ - 2x)] = \sqrt{3}[\sin(60^\circ + 2x) + \sin(60^\circ - 2x)]

Step 2: Use the sine addition identities
From previous steps, we have:

sin(60+2x)+sin(602x)=2sin(60)cos(2x)\sin(60^\circ + 2x) + \sin(60^\circ - 2x) = 2 \sin(60^\circ) \cos(2x)

Thus:

Step 3: Substitute for sine
Using (\sin 60^\circ = \frac{\sqrt{3}}{2}), we can now write:

2332cos(2x)=3cos(2x).2 \sqrt{3} \cdot \frac{\sqrt{3}}{2} \cos(2x) = 3 \cos(2x).

Step 4: Answer in terms of t
Recalling (\cos^2 x = 1 - \sin^2 x = 1 - (1 - t) = t), we derive:

=3[t211].= 3[\sqrt{t^2 - 1} - 1].

Thus, the final expression gives:

6t3\boxed{\frac{6}{t} - 3}.

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