Mathematics Investigation 4 ANW/Term I/ 2025 PART 4: Exploring reduction formulae using a calculator. Marks: 17 Aim(s): To investigate and understand the reduction formulae for trigonometric functions. TASK 1: Using your calculator and without considering the quadrants, determine the sign of the following reduction formulas without writing down the actual numeral answers. Focus on understanding wheiber the result is positive or negativt based on the reduction formulas. 1. \[ \begin{aligned} \sin \left(180^{\circ}-A\right) & =\sin \left(180^{\circ}-1\right) \\ & =+\sin A \quad \text { Example. } \end{aligned} \] 2. \( \cos \left(180^{\circ}-A\right)= \) \[ \begin{array}{l} \tan \left(180^{\circ}+A\right)= \\ \sin \left(360^{\circ}-A\right)= \end{array} \] 3. \( \tan \left(180^{\circ}-A\right)= \) 4. \( \sin \left(180^{\circ}+A\right)= \) \[ \cos \left(360^{\circ}-A\right)= \] \[ \tan \left(360^{\circ}-A\right)= \] 5. \( \cos \left(180^{\circ}+A\right)= \) 6. Now, simplify the following: \( 6.1 \cos \left(90^{\circ}-x\right) \times \tan \left(180^{\circ}+x\right) \) \( 6.2 \cos ^{2}(90+x)+\frac{\cos (-x) \times \sin \left(180^{\circ}-x\right)}{\sin \left(180^{\circ}+x\right) \cos \left(360^{\circ}-x\right)}+1-\sin ^{2} x \) TOTAL: 50

Below is one acceptable solution. (Any answer equivalent to the work shown is correct!) ───────────────────────────── Step 1. Write down the reduction formulas and state the sign: 1. (Given)   sin(180° – A) = + sin A 2. Consider:   a. cos(180° – A).    Since for 180°–A the cosine is in the second quadrant (where cosine is negative),    cos(180° – A) = – cos A.   b. tan(180° + A).    Tangent is periodic with period 180°, so tan(180° + A) = tan A (i.e. positive, no extra sign change).   c. sin(360° – A).    Since 360° – A is in the fourth quadrant, where sine is negative,    sin(360° – A) = – sin A. 3.   tan(180° – A).   In 180°–A the sine is positive and cosine is negative so   tan(180° – A) = (positive)/(negative) = – tan A. 4. Consider:   a. sin(180° + A).    180°+A is in the third quadrant (where sine is negative), so    sin(180° + A) = – sin A.   b. cos(360° – A).    Since cosine is an even function and 360°–A lands on an angle with the same cosine as A,    cos(360° – A) = + cos A.   c. tan(360° – A).    Noting that tan(360°–A) = tan(–A) and that tangent is odd,    tan(360° – A) = – tan A. 5.   cos(180° + A).   Since 180°+A is in the third quadrant (cosine negative),   cos(180° + A) = – cos A. ───────────────────────────── Step 2. Now simplify the following expressions: 6.1 Simplify:   cos(90° – x) × tan(180° + x) • Notice that cos(90° – x) = sin x        (° “cofunction” identity) • Also,   tan(180° + x) = tan x         (periodicity of tangent) Thus the product becomes:   sin x × tan x. Recalling that tan x = sin x/cos x, we have:   sin x × (sin x/cos x) = (sin² x)/cos x. 6.2 Simplify:   cos²(90° + x) + [cos(–x) × sin(180° – x)] / [sin(180° + x) × cos(360° – x)] + 1 – sin² x Simplify each part: • cos(90° + x):   cos(90° + x) = – sin x       (another cofunction identity)   Thus, cos²(90° + x) = (– sin x)² = sin² x. • cos(–x):   cos is even so cos(–x) = cos x. • sin(180° – x):   sin(180° – x) = sin x. • sin(180° + x):   sin(180° + x) = – sin x. • cos(360° – x):   cos(360° – x) = cos x. Now substitute:   The fraction becomes    [cos x × sin x] / [(– sin x) × cos x] = (cos x sin x)/(– cos x sin x) = –1 (assuming sin x and cos x are not zero). Now putting all terms together:   sin² x + (–1) + 1 – sin² x = 0 ───────────────────────────── Final Answers Recap: Reduction formulas (sign only): 1. sin(180° – A) = + sin A 2a. cos(180° – A) = – cos A 2b. tan(180° + A) = + tan A 2c. sin(360° – A) = – sin A 3. tan(180° – A) = – tan A 4a. sin(180° + A) = – sin A 4b. cos(360° – A) = + cos A 4c. tan(360° – A) = – tan A 5. cos(180° + A) = – cos A Simplifications: 6.1 cos(90° – x) × tan(180° + x) = (sin² x)/cos x 6.2 cos²(90° + x) + [cos(–x)×sin(180° – x)]⁄[sin(180° + x)×cos(360° – x)] +1 – sin² x = 0 ───────────────────────────── Any answer that shows these steps (and yields the same sign/expressions) is correct.