5.1 Simplify (WITHOUT THE USE OF A CALCULATOR) sin(180° - 360°) . cos(x) . tan(180° + x) . cos(-x) tan(-x) . cos(90° - x) . sin(90° - x) 5.2 Prove the identity: \[ \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x} = \frac{2}{\sin x} \] 5.3 Use compound angles to show that: \[ \cos 2x = 2\cos^2 x - 1 \] 5.4 Determine the general solution for $ x $ if: \[ \cos 2x + 3 = 2 \] 5.5 In $ \triangle ABC: \; A + B = 90° $. Determine the value of $ \sin A \cdot \cos B + \cos A \cdot \sin B $.

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5.1 Simplify (WITHOUT THE USE OF A CALCULATOR) sin(180° - 360°) - NSC Mathematics - Question 5 - 2016 - Paper 2

Question 5

5.1 Simplify (WITHOUT THE USE OF A CALCULATOR) sin(180° - 360°) . cos(x) . tan(180° + x) . cos(-x) tan(-x) . cos(90° - x) . sin(90° - x) 5.2 Prove the identity: \... show full transcript

Worked Solution & Example Answer:5.1 Simplify (WITHOUT THE USE OF A CALCULATOR) sin(180° - 360°) - NSC Mathematics - Question 5 - 2016 - Paper 2

Step 1

5.1 Simplify (WITHOUT THE USE OF A CALCULATOR)

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Answer

To simplify the expression:

Start with:

ext{Expression} = ext{sin}(180° - 360°) imes ext{cos}(x) imes ext{tan}(180° + x) imes ext{cos}(-x) \ ext{tan}(-x) imes ext{cos}(90° - x) imes ext{sin}(90° - x)

Using trigonometric identities:

\ ext{sin}(180° - θ) = ext{sin} θ
\ ext{cos}(180° + θ) = - ext{cos} θ
\ ext{cos}(-θ) = ext{cos} θ
\ ext{tan}(-θ) = - ext{tan} θ
\ ext{cos}(90° - θ) = ext{sin} θ

Apply these identities to the expression:

= ext{sin}(360°) imes ext{cos}(x) imes (- ext{tan}(x)) imes ext{cos}(x) imes (- ext{tan}(x)) imes ext{sin}(x) imes ext{sin}(x)

With sin(360°) = 0, the entire expression simplifies to:

$ext{Expression} = - ext{cos}(x)$

Step 2

5.2 Prove the identity:

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Answer

To prove the identity:

Start with the left-hand side (LHS):

LHS = \frac{\sin x}{1 + \cos x} + \frac{1 + \cos x}{\sin x}

Finding a common denominator:

LHS = \frac{\sin^2 x + (1 + \cos x)(1 + \cos x)}{\sin x (1 + \cos x)}\ = \frac{\sin^2 x + 1 + 2\cos x + \cos^2 x}{\sin x (1 + \cos x)}

Now using the Pythagorean identity, (\sin^2 x + \cos^2 x = 1):

= \frac{2 + 2\cos x}{\sin x(1 + \cos x)} = \frac{2}{\sin x}

Thus, (LHS = RHS), confirming the identity.

Step 3

5.3 Use compound angles to show that:

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Answer

To show that: $\cos 2x = 2\cos^2 x - 1$

Using the compound angle formula for cosine:

We can express (\sin^2 x) in terms of (\cos^2 x): $\sin^2 x = 1 - \cos^2 x$

Substituting this into the equation: $\cos 2x = \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1$

This confirms the identity.

Step 4

5.4 Determine the general solution for x if:

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Answer

To solve:

Rearranging gives:

The general solutions for (\cos \theta = -1) are:

Thus:

Step 5

5.5 In $ \triangle ABC: \; A + B = 90° $. Determine the value of $ \sin A \cdot \cos B + \cos A \cdot \sin B $.

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Answer

Using the identity for sine of a sum:

Since (A + B = 90°), this implies:

Thus:

5.1 Simplify (WITHOUT THE USE OF A CALCULATOR) sin(180° - 360°) - NSC Mathematics - Question 5 - 2016 - Paper 2

Worked Solution & Example Answer:5.1 Simplify (WITHOUT THE USE OF A CALCULATOR) sin(180° - 360°) - NSC Mathematics - Question 5 - 2016 - Paper 2

5.1 Simplify (WITHOUT THE USE OF A CALCULATOR)

5.2 Prove the identity:

5.3 Use compound angles to show that:

5.4 Determine the general solution for x if:

5.5 In \( \triangle ABC: \; A + B = 90° \). Determine the value of \( \sin A \cdot \cos B + \cos A \cdot \sin B \).

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