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Lecture Notes Simplifying Trigonometric Expressions page 1 Sample Problems Find the exact value of each of the following and simplify. 1. 2 sin 60 5 tan 30 cot 45 2. 2 sin 60 3 cos 45 + tan 30 3. 4 cos 30 sin 45 + tan 60 4. sin 30 cos 45 tan 60 5. 2 sec 60 +3 csc 45 tan 30 cos 60 sin 60 1 cos 30 + 1 Practice Problems Find the exact value of each of the following and simplify. 1. sin 45 2. 3 tan 60 3. 2 cos 60 tan 30 csc 45 3 cos 45 sec 60 cos 45 tan2 60 5. 3 sec 30 cos 60 sin 60 tan 60 tan 30 tan 60 + tan 30 7. sin 30 cos 30 tan 30 cot 30 sin 45 cos 45 tan 45 cot 45 tan 30 tan 60 csc 45 cos 45 cot 45 sin 45 + tan 45 4. sin2 30 6. 8. 2 cot 60 tan 60 c copyright Hidegkuti, Powell, 2012 9. 1 2 tan 30 tan2 30 (sin 30 ) (cos 30 ) (sin 45 ) (cos 45 ) (tan 30 ) cot 30 (tan 45 ) cot 45 Last revised: September 27, 2012 Simplifying Trigonometric Expressions Lecture Notes page 2 Sample Problems - Answers p 2 3 3 1.) 3p 6 4.) 4 2.) p 8 3 p 9 2 p 4 3 or 3 6 p 3 6 16 4 or 4 5.) p 3 2 2 p 3.) 3 3 p p p 2 6 3 2 or 2 2 p 7+4 3 Practice Problems - Answers p 2 2 1.) 8.) p 3 p 3 3 9.) 2 p 2.) 3 3 p 4 2 p 3.) 2 2 3 4.) p 3 2 2 7 4 7p 5.) 3 4 1 6.) 2 7.) p 3 2 p 3 2 c copyright Hidegkuti, Powell, 2012 Last revised: September 27, 2012 Simplifying Trigonometric Expressions Lecture Notes page 3 Sample Problems - Solutions Find the exact value of each of the following and simplify. p 2 3 1. 2 sin 60 5 tan 30 cot 45 = 3 Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles, we obtain that p 3 1 sin 60 = tan 30 = p and cot 45 = 1 2 3 Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst. p p 3 3 1 tan 30 = p p = 3 3 3 p p p p 3 3 5 3 5 1= 3 2 sin 60 5 tan 30 cot 45 = 2 2 3 3 We will then bring these to the common denominator p 2. 2 sin 60 3 p p 5 3 3 3 = 3 3 p p p 5 3 3 3 5 3 = = 3 3 p 2 3 3 3 cos 45 + tan 30 Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles, we obtain that p 3 1 1 cos 45 = p and tan 30 = p sin 60 = 2 2 3 Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst. p p p p 3 3 2 1 2 1 and tan 30 = p p = cos 45 = p p = 3 2 3 3 2 2 p p p p p 3 3 2 2 3 p 3 + 2 sin 60 3 cos 45 + tan 30 = 2 3 + = 3 2 3 2 2 3 We will then bring these to the common denominator, which is 6 and if there are like terms, we combine them. p p p p p p 6 3 9 2 2 3 3 3 2 = + 3 + 3 6 6 6 2 p p p p p 6 3 9 2+2 3 8 3 9 2 = = 6 6 Note: p 8 3 3. 4 cos 30 p 9 2 6 p 8 3 = 6 p p 9 2 4 3 = 6 3 p 3 2 is also perfectly acceptable as …nal answer. 2 sin 45 + tan 60 Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles, we obtain that p p 1 3 cos 30 = sin 45 = p and tan 60 = 3 2 2 c copyright Hidegkuti, Powell, 2012 Last revised: September 27, 2012 Lecture Notes Simplifying Trigonometric Expressions page 4 Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst. p p 1 2 2 sin 45 = p p = 2 2 2 p p p p p p 3 2 p 2 2 p 4 cos 30 sin 45 + tan 60 = 4 + 3=2 3 + 3= 3 3 2 2 2 2 Note: We can also bring these to the common denominator. p p p p p p 2 6 3 2 6 3 2 3 3 = = 2 2 2 2 is also perfectly acceptable. 3p 6 4 4 Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles, we obtain that p p 1 1 1 1 sin 30 = cos 45 = p tan 60 = 3 sec 60 = 2 csc 45 = 2 tan 30 = p and cos 60 = 2 2 3 2 4. sin 30 cos 45 tan 60 2 sec 60 + 3 csc 45 tan 30 cos 60 = Because we will be adding and subtracting these expressions, it is best if we rationalize them …rst. p p 2 3 and tan 30 = cos 45 = 2 3 sin 30 cos 45 tan 60 2 sec 60 + 3 csc 45 tan 30 cos 60 = p p p p p p p p p p 3 1 6 3 6 6 6 6 2 6 3 6 1 2 p 3 2 2+3 2 = 4+ = 4+ = 4+ = = 6 4 2 4 4 4 3 2 4 2 2 Note: We can also bring these to the common denominator. p p p 3 6 3 6 16 3 6 16 4= = 4 4 4 4 4 is also perfectly acceptable. p sin 60 1 5. = 7+4 3 cos 30 + 1 Solution: We …rst obtain the exact values of the trigonometric expressions. Using the famous right triangles, we obtain that p 3 p p 1 sin 60 1 3 3 sin 60 = and cos 30 = and so = p2 2 2 cos 30 + 1 3 +1 2 To clear the denominators, we multiply both numerator and denominator by 2 ! p p 3 3 2 1 p 1 2 3 2 !=p p2 = p 3+2 3 3 +1 2 +1 2 2 We rationalize our result: p p p 3 2 3 2 3 p p =p 3+2 3+2 3 2 = 2 p 3 3 2 4 2 = 3 p 4 3+4 7 = 1 p 4 3 = 1 7 p 4 3 = p 7+4 3 For more documents like this, visit our page at http://www.teaching.martahidegkuti.com and click on Lecture Notes. E-mail questions or comments to [email protected]. c copyright Hidegkuti, Powell, 2012 Last revised: September 27, 2012