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\( \begin{array}{ll}\text { (3) } \frac{\cos ^{2} 165^{\circ}}{\cos 60^{\circ}}-\cos ^{2} 97^{\circ}-\cos ^{2} 727^{\circ} & \text { (4) } \frac{\cos 204^{\circ} \tan 315^{\circ}+2 \cos 156^{\circ} \sin ^{2} 348}{\sin ^{2}\left(-5^{\circ}\right)+\sin ^{2} 85^{\circ}-\sin ^{2} 24^{\circ}} \\ \text { (5) } \sin 140^{\circ}-\cos 220^{\circ} \tan \left(-155^{\circ}\right) & \text { (6) } \frac{\cos ^{4} 190^{\circ}-\sin ^{4}\left(-10^{\circ}\right)}{\cos 55^{\circ} \cos 325^{\circ}}\end{array} \)

Ask by Weaver Peterson. in South Africa
Feb 28,2025

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The answers are: (3) (√3)/2 (4) 1 (5) 1 (6) 2

Solución

We will simplify each of the four expressions step‐by‐step. ───────────────────────────── Step (3): Evaluate   [cos²(165°)]⁄[cos(60°)] – cos²(97°) – cos²(727°). 1. Notice that cos 60° = ½ so   cos²(165°)/cos60° = cos²(165°)/(½) = 2·cos²(165°). 2. Write the angles in terms of acute angles:  • 165° = 180° – 15° ⟹ cos(165°) = –cos15° ⟹ cos²(165°) = cos²15°.  • 97° = 90° + 7°  ⟹ cos(97°) = –sin7° ⟹ cos²(97°) = sin²7°.  • 727° differs by full rotations. Since 727° – 720° = 7°, we have   cos²(727°) = cos²7°. 3. The expression becomes:   2·cos²15° – sin²7° – cos²7°.  But recall that sin²7° + cos²7° = 1.  Thus, the expression is 2·cos²15° – 1. 4. Recognize the double‐angle formula:   2·cos²15° – 1 = cos30°.  Since cos30° = (√3)/2, the final answer for (3) is (√3)/2. ───────────────────────────── Step (4): Evaluate   [cos(204°)·tan(315°) + 2·cos(156°)·sin²(348°)] ÷ [sin²(–5°) + sin²(85°) – sin²(24°)]. Numerator: 1. For cos(204°): 204° = 180° + 24° so cos(204°) = –cos24°. 2. For tan(315°): 315° = 360° – 45° so tan(315°) = –tan45° = –1.  Then, cos(204°)·tan(315°) = (–cos24°)(–1) = cos24°. 3. For cos(156°): 156° = 180° – 24° so cos(156°) = –cos24°. 4. For sin(348°): 348° = 360° – 12° so sin(348°) = –sin12°, and sin²(348°) = sin²12°.  Then, 2·cos(156°)·sin²(348°) = 2·(–cos24°)·sin²12° = –2·cos24°·sin²12°. 5. The entire numerator becomes:   cos24° – 2·cos24°·sin²12° = cos24°·(1 – 2 sin²12°). 6. Recognize that 1 – 2 sin²x = cos(2x). Here, with x = 12°, we have:   1 – 2 sin²12° = cos24°.  Thus the numerator equals:   cos24° · cos24° = cos²24°. Denominator: 1. Note that sin²(–5°) = sin²5° because sine squared is even. 2. For sin(85°): 85° = 90° – 5° so sin85° = cos5°, and sin²85° = cos²5°. 3. Then, the denominator is:   sin²5° + cos²5° – sin²24° = 1 – sin²24°,  since sin²5° + cos²5° = 1. 4. But 1 – sin²24° = cos²24°. Thus, the entire fraction is:   cos²24° / cos²24° = 1. ───────────────────────────── Step (5): Evaluate   sin(140°) – cos(220°)·tan(–155°). 1. Write the angles in familiar forms:  • 140° = 180° – 40° ⟹ sin140° = sin40°.  • 220° = 180° + 40° ⟹ cos220° = –cos40°.  • tan(–155°) = –tan155° (since tan is odd).  • Also, 155° = 180° – 25° ⟹ tan155° = –tan25°, so   tan(–155°) = –(–tan25°) = tan25°. 2. Substitute these results back:   sin(140°) – cos(220°)·tan(–155°) = sin40° – [–cos40°]·tan25° = sin40° + cos40°·tan25°. 3. Write tan25° as sin25°/cos25°:   This becomes sin40° + (cos40°·sin25°)/cos25°. 4. Combine over a common denominator:   = [sin40°·cos25° + cos40°·sin25°] / cos25°.  The numerator is recognizable as sin(40° + 25°) by the sine addition formula:   sin40°·cos25° + cos40°·sin25° = sin65°. 5. Also note cos25° = sin(90°–25°) = sin65°.  Thus the fraction is sin65° / sin65° = 1. ───────────────────────────── Step (6): Evaluate   [cos⁴(190°) – sin⁴(–10°)] ÷ [cos(55°)·cos(325°)]. 1. Simplify the numerator:  • 190° = 180° + 10° ⟹ cos190° = –cos10° so cos⁴190° = (–cos10°)⁴ = cos⁴10°.  • sin(–10°) = –sin10° so sin⁴(–10°) = (–sin10°)⁴ = sin⁴10°.  Thus, the numerator is cos⁴10° – sin⁴10°. 2. Factor the difference of two fourth powers as a difference of squares:   cos⁴10° – sin⁴10° = (cos²10°)² – (sin²10°)² = (cos²10° – sin²10°)(cos²10° + sin²10°).  Since cos²10° + sin²10° = 1, this simplifies to:   cos²10° – sin²10°.  But cos²10° – sin²10° = cos20° (by the double‐angle formula). 3. Now simplify the denominator:  • 325° = 360° – 35° ⟹ cos325° = cos35°.  So the denominator is cos55°·cos35°.  Using the product-to-sum formula:   cos55°·cos35° = [cos(55° + 35°) + cos(55° – 35°)]/2 = [cos90° + cos20°]/2 = [0 + cos20°]/2 = cos20°/2. 4. Thus the whole expression becomes:   cos20° ÷ (cos20°/2) = 2. ───────────────────────────── Final Answers:  (3) = (√3)/2  (4) = 1  (5) = 1  (6) = 2

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Extra Insights

Did you know that the cosine and sine functions have fascinating historical ties to ancient mathematics? The study of trigonometric functions dates back to ancient civilizations, notably the Babylonians and Greeks, who used them in astronomy. They initially described these functions in terms of chords in a circle, paving the way for the modern sine and cosine that we use in mathematics today! If you're diving into solving those trigonometric expressions, remember to always simplify angles first! Angles can exceed 360 degrees or fall below 0, but using the periodic properties (like 360° for sine and cosine) can make your life so much easier. Additionally, pay attention to the signs of functions based on the quadrant the angle lies in—this can be a common pitfall for many. Happy calculating!

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