Use Example 1.10 below to find all six trigonometric functions of 15? Example 1.10 Find the sine,

Use Example 1.10 below to find all six trigonometric functions of 15?
 

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Example 1.10 Find the sine, cosine, and tangent of 75°. D Solution: Since 75° = 45°+30°, place a 30-60-90 right triangle AADB with legs of length v3 and 1 on top of the hypotenuse of a 45 - 45- 90 right triangle AABC whose hypotenuse has length v3, as in the figure on the right. From Figure 1.2.2(a) we know that the length of each leg of AABC is the length of the . Draw DE 1 B hypotenuse divided by v2. So AC = BC = perpendicular to AC, so that AADE is a right triangle. Since ZBAC = 45° and ZDAB = 30°, we see that ZDAE = 75° since it is the sum of those two angles. Thus, we need to find the sine, cosine, and tangent of ZDAE. Notice that ZADE = 15°, since it is the complement of ZDAE. And ZADB = 60°, since it is the complement of ZDAB. Draw BF perpendicular to DE, so that ADFB is a right triangle. 2 Then ZBDF = 45°, since it is the difference of ZADB = 60° and ZADE = 15°. Also, ZDBF = 45° since it is the complement of ZBDF. The hypotenuse BD of ADFB has length 1 and ADFB is a 45– 45– 90 right triangle, so we know that DF = FB =. Now, we know that DE 1 AC and BCIAC, so FE and BC are parallel. Likewise, FB and EC are both perpendicular to DE and hence FB is parallel to EC. Thus, FBCE is a rectangle, since ZBCE is a right angle. So EC = FB = and FE = BC = V3. Hence, 45° A DE = DF + FE = + + V = V3 + 1 V3-1. Thus, and AE = AC - EC = 3+1 sin 75° = AD = V6rv2 , cos 75° = A == v v2 , and tan 75° = == + %3D !! V6-V2 Note: Taking reciprocals, we get csc 75° = sec 75° = , and cot 75° = V6-V2 V6+v2 |rolen o08 Example 1.10 Find the sine, cosine, and tangent of 75°. D Solution: Since 75° = 45°+30°, place a 30-60-90 right triangle AADB with legs of length v3 and 1 on top of the hypotenuse of a 45 - 45- 90 right triangle AABC whose hypotenuse has length v3, as in the figure on the right. From Figure 1.2.2(a) we know that the length of each leg of AABC is the length of the . Draw DE 1 B hypotenuse divided by v2. So AC = BC = perpendicular to AC, so that AADE is a right triangle. Since ZBAC = 45° and ZDAB = 30°, we see that ZDAE = 75° since it is the sum of those two angles. Thus, we need to find the sine, cosine, and tangent of ZDAE. Notice that ZADE = 15°, since it is the complement of ZDAE. And ZADB = 60°, since it is the complement of ZDAB. Draw BF perpendicular to DE, so that ADFB is a right triangle. 2 Then ZBDF = 45°, since it is the difference of ZADB = 60° and ZADE = 15°. Also, ZDBF = 45° since it is the complement of ZBDF. The hypotenuse BD of ADFB has length 1 and ADFB is a 45– 45– 90 right triangle, so we know that DF = FB =. Now, we know that DE 1 AC and BCIAC, so FE and BC are parallel. Likewise, FB and EC are both perpendicular to DE and hence FB is parallel to EC. Thus, FBCE is a rectangle, since ZBCE is a right angle. So EC = FB = and FE = BC = V3. Hence, 45° A DE = DF + FE = + + V = V3 + 1 V3-1. Thus, and AE = AC - EC = 3+1 sin 75° = AD = V6rv2 , cos 75° = A == v v2 , and tan 75° = == + %3D !! V6-V2 Note: Taking reciprocals, we get csc 75° = sec 75° = , and cot 75° = V6-V2 V6+v2 |rolen o08

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Answers Since ADFB is an isoceles right triangle ... View the full answer