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Trigonometry Calculations and Solutions

Solve trigonometry problems without a calculator, finding exact values and decimal approximations, and explore various quadrants and trigonometric identities.

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Trigonometry Calculations and Solutions

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  1. 1. Without using your calculator, find the exact value of each of the following: (i) tan 180° 90° 0° 180° 360° S A 270° C T

  2. 1. Without using your calculator, find the exact value of each of the following: (ii) sin 135° 90° 2nd quadrant  sin is positive sin 135° = + sin (180° – 135°) 0° 180° 360° = sin 45° S A 270° C T

  3. 1. Without using your calculator, find the exact value of each of the following: (iii) cos 210° 90° 3rd quadrant  cos is negative cos 210° = – cos(210° – 180°) 0° 180° 360° = – cos 30° S A 270° C T

  4. 1. Without using your calculator, find the exact value of each of the following: (iv) cos 315° 90° 4th quadrant  cos is positive cos 315° = + cos(360° – 315°) 0° 180° 360° = cos 45° S A 270° C T

  5. 1. Without using your calculator, find the exact value of each of the following: (v) sin 240° 90° 3rd quadrant  sin is negative sin 240° = – sin(240° – 180°) 0° 180° 360° = – sin 60° S A 270° C T

  6. 2. Showing all your work clearly, find the value of the following, correct to four places of decimal. 90° (i) sin 295° 4th quadrant  sin is negative S A sin 295° = – sin (360° – 295°) 180° 0° 360° C T = – sin 65° = – 0·9063 295° 270°

  7. 2. Showing all your work clearly, find the value of the following, correct to four places of decimal. 110° 90° (ii) cos 110° 2nd quadrant  cos is negative S A cos 110° = – cos(180° – 110°) 180° 0° 360° C T = – cos 70° = – 0·3420 270°

  8. 2. Showing all your work clearly, find the value of the following, correct to four places of decimal. 90° (iii) tan 230° 3rd quadrant  tan is positive S A 180° 0° tan 230° = + tan (230° – 180°) 360° C T = tan 50° = 1·19175 = 1·1918 230° 270°

  9. 3. (i) Showing all your work clearly, find sin 160°, correct to four places of decimal. 90° sin 160° 2nd quadrant  sin is positive 160° S A sin 160° = + sin (180° – 160°) 180° 0° 360° C T = sin 20° = 0·3420 270°

  10. 3. (ii) Showing all your work clearly, find cos 250°, correct to four places of decimal. 90° cos 250° 3rd quadrant  cos is negative S A 180° cos 250° = – cos(250° – 180º) 0° 360° C T = – cos 70° = – 0·3420 270°

  11. 3. (iii) Hence show that sin 160° – cos 250° > 0. sin 160° – cos 250° = 0·3420 – (– 0·3420) = 0·3420 + 0·3420 = 0·6840 > 0 Therefore, sin 160° – cos 250° > 0

  12. 4. Given that sin θ, solve for all possible values of θ for 0° ≤θ≤ 360°. 90° 135° 45° sin is positive in the 1st and 2nd quadrant Reference angle S A 0° 180° 360° C T 1st quadrant: 45º 2nd quadrant: 180º – 45º = 135º 270°

  13. 5. Given that cosθ, solve for all possible values of θ for 0° ≤θ≤ 360°. 90° 120° cos is negative in the 2nd and 3rd quadrant Reference angle S A 0° 180° 360° C T = 60° 240° 2nd Quadrant:180° – 60° = 120° 270° 3rd Quadrant:180° + 60° = 240°

  14. 6. Given that tan θ= -1, solve for all possible values ofθfor 0°  360°. 90° 135° tan is negative in the 2nd and 4th quadrants Reference angle 0° tan θ = 1 180° 360° θ = tan–1 1 S A = 45° C T 315° 2nd Quadrant:180° – 45° = 135° 270° 4th Quadrant:360° – 45° = 315°

  15. 7. Find, correct to the nearest degree, the two values of θ for 0°  360°, given that (i) sin θ = – 0·9744 90° sin is negative in the 3rd and 4th quadrant. Reference angle sin θ = 0·9744 S A 0° θ = sin–1 0·9744 180° 360° C T = 77° 3rd quadrant = 180° + 77° = 257° 4th quadrant = 360° – 77° = 283° 257° 283° 270°

  16. 7. Find, correct to the nearest degree, the two values of θ for 0°  360°, given that (ii) tan θ = 1·28 90° tan is positive in the 1st and 3rd quadrant 52° Reference angle tan θ = 1·28 S A 0° θ = tan–1 1·28 180° 360° C T θ = 52° 1st quadrant = 52° 232° 270° 3rd quadrant = 180° + 52° = 232°

  17. 7. Find, correct to the nearest degree, the two values of θ for 0°  360°, given that (iii) cos θ = – 0·3748 112° cos is negative in the 2nd and 3rd quadrant Reference angle S A cosθ = 0·3748 C T θ = cos–1 0·3745 θ = 68° 248° 2nd Quadrant: 180° – 68 = 112° 3rdQuadrant= 180° + 68 = 248°

  18. 8. Given that sin A = and cos A is positive, find the value of A in the range 0° A  360°. cosA is positive sin negative, cos positive  4th quadrant S A Reference angle C T = 60° 4th quadrant = 360° – 60° = 300°

  19. 9. Given that tan B = and sin B is negative, find the value of B in the range 0° B  360°. sin B is negative tan positive, sin negative  3rd quadrant S A Reference angle C T B = 30° 3rd quadrant = 180° + 30° = 210°

  20. 10. If and 90° A 180°, without finding A, find the values of tan A and cos A. (Hyp) (Opp) Find the adjacent side: 52 + (Adj)2 = 132 25 + (Adj)2 = 169 (Adj)2 = 169 – 25 (Adj)2 = 144 Adj = Adj = 12

  21. 10. If and 90° A 180°, without finding A, find the values of tan A and cos. (Hyp) (Opp) S A 12 But 90° A  180° C T  2nd quadrant  cos and tan are negative

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