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(d) Calculate each of the following without the use of a calculator: (1) \( \cos 80^{\circ} \cos 10^{\circ}-\sin 80^{\circ} \sin 10^{\circ} \) (2) \( \cos 70^{\circ} \cos 40^{\circ}+\sin 70^{\circ} \sin 40^{\circ} \) (3) \( \sin 280^{\circ} \cos 160^{\circ}-\cos 100^{\circ} \sin 200^{\circ} \) (4) \( \cos 265^{\circ} \sin 355^{\circ}-\sin 85^{\circ} \cos 175^{\circ} \) (5) \( \cos 65^{\circ} \cos 295^{\circ}-\sin 115^{\circ} \cos 205^{\circ} \) (6) \( \cos 50^{\circ} \sin 260^{\circ}+\cos 10^{\circ} \sin 140^{\circ} \)

Ask by Johnston Hampton. in South Africa
Feb 20,2025

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1. 0 2. √3/2 3. √3/2 4. 1 5. 1 6. -√3/2

Solución

To solve each of the trigonometric expressions, we will use the appropriate trigonometric identities. ### (1) \( \cos 80^{\circ} \cos 10^{\circ} - \sin 80^{\circ} \sin 10^{\circ} \) This expression can be simplified using the cosine of a sum identity: \[ \cos(A) \cos(B) - \sin(A) \sin(B) = \cos(A + B) \] Here, \( A = 80^{\circ} \) and \( B = 10^{\circ} \): \[ \cos(80^{\circ}) \cos(10^{\circ}) - \sin(80^{\circ}) \sin(10^{\circ}) = \cos(80^{\circ} + 10^{\circ}) = \cos(90^{\circ}) = 0 \] ### (2) \( \cos 70^{\circ} \cos 40^{\circ} + \sin 70^{\circ} \sin 40^{\circ} \) This expression can be simplified using the cosine of a difference identity: \[ \cos(A) \cos(B) + \sin(A) \sin(B) = \cos(A - B) \] Here, \( A = 70^{\circ} \) and \( B = 40^{\circ} \): \[ \cos(70^{\circ}) \cos(40^{\circ}) + \sin(70^{\circ}) \sin(40^{\circ}) = \cos(70^{\circ} - 40^{\circ}) = \cos(30^{\circ}) = \frac{\sqrt{3}}{2} \] ### (3) \( \sin 280^{\circ} \cos 160^{\circ} - \cos 100^{\circ} \sin 200^{\circ} \) This expression can be simplified using the sine of a difference identity: \[ \sin(A) \cos(B) - \cos(A) \sin(B) = \sin(A - B) \] Here, \( A = 280^{\circ} \) and \( B = 160^{\circ} \): \[ \sin(280^{\circ}) \cos(160^{\circ}) - \cos(100^{\circ}) \sin(200^{\circ}) = \sin(280^{\circ} - 160^{\circ}) = \sin(120^{\circ}) = \frac{\sqrt{3}}{2} \] ### (4) \( \cos 265^{\circ} \sin 355^{\circ} - \sin 85^{\circ} \cos 175^{\circ} \) Using the sine of a difference identity: \[ \sin(A) \cos(B) - \cos(A) \sin(B) = \sin(A - B) \] Here, \( A = 355^{\circ} \) and \( B = 265^{\circ} \): \[ \cos(265^{\circ}) \sin(355^{\circ}) - \sin(85^{\circ}) \cos(175^{\circ}) = \sin(355^{\circ} - 265^{\circ}) = \sin(90^{\circ}) = 1 \] ### (5) \( \cos 65^{\circ} \cos 295^{\circ} - \sin 115^{\circ} \cos 205^{\circ} \) Using the cosine of a sum identity: \[ \cos(A) \cos(B) - \sin(A) \sin(B) = \cos(A + B) \] Here, \( A = 65^{\circ} \) and \( B = 295^{\circ} \): \[ \cos(65^{\circ}) \cos(295^{\circ}) - \sin(115^{\circ}) \cos(205^{\circ}) = \cos(65^{\circ} + 295^{\circ}) = \cos(360^{\circ}) = 1 \] ### (6) \( \cos 50^{\circ} \sin 260^{\circ} + \cos 10^{\circ} \sin 140^{\circ} \) Using the sine of a sum identity: \[ \sin(A) \cos(B) + \cos(A) \sin(B) = \sin(A + B) \] Here, \( A = 260^{\circ} \) and \( B = 50^{\circ} \): \[ \cos(50^{\circ}) \sin(260^{\circ}) + \cos(10^{\circ}) \sin(140^{\circ}) = \sin(260^{\circ} + 50^{\circ}) = \sin(310^{\circ}) = -\frac{\sqrt{3}}{2} \] ### Summary of Results: 1. \( 0 \) 2. \( \frac{\sqrt{3}}{2} \) 3. \( \frac{\sqrt{3}}{2} \) 4. \( 1 \) 5. \( 1 \) 6. \( -\frac{\sqrt{3}}{2} \)

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Bonus Knowledge

When diving into trigonometric identities, it's always fun to remember their historical roots! The foundation of trigonometry dates back to ancient civilizations like the Greeks and Indians, where they explored the relationships of angles and sides in triangles. The sine and cosine functions were first formalized in India around the 7th century, showcasing not just a mathematical advancement but a cultural one that influenced astronomy and navigation! As you tackle these calculations, keep in mind common pitfalls. One well-known mistake is misapplying angle identities, like mixing up sine and cosine values when using co-function identities (e.g., \(\sin(90^\circ - x) = \cos(x)\)). Additionally, remembering the key angles—like \( 0^\circ, 30^\circ, 45^\circ, 60^\circ, \) and \( 90^\circ \)—can save you from unnecessary confusion and help you calculate more efficiently. Happy calculating!

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