Trigonometric functions of sharp angle Trigonometry was originally an area of mathematics that deal with calculating unknown elements of the triangle with well-known. The name comes from two Greek words , TRIGONOS-meaning triangle and METRON-meaning measures. How do we define trigonometric functions? Observe right-angled triangle ABC. a, b → cathetus c → hypotenuse a2 + b2 = c2 → Pythagorean theorem opposite a sinα = = hypotenuse c adjacent a cosα = = hypotenuse c opposite a tgα = = adjacent b adjacent b ctgα = = opposite a Take heed: The symbol sin (cos, tg, ctg) will not indicate any size! Always must have angle! How to calculate values of trigonometric functions for the angles 30o ,45o and 60o ? a opposite a 1 sin 30o = = 2 = = hypotenuse a 2a 2 a 3 adjacent 3 cos30o = = 2 = hypotenuse a 2 a opposite 1 1 3 3 tg30o = = 2 = = ⋅ = adjacent 3 3 3 3 3 2 a 3 adjacent ctg30o = = 2 = 3 opposite a 2 Now we'll do (by definition) for angle 60o : a 3 3 sin 60o = 2 = a 2 a 1 cos60o = 2 = a 2 3 3 tg60o = 2 = a 3 2 a 3 ctg60o = 2 = a 3 3 2 For the value of the trigonometric functions of angle 45o , we will use half of the square. opposite a 1 2 2 sin 45o = = = ⋅ = hypotenuse a 2 2 2 2 adjacent a 2 cos 45o = = = hypotenuse 2 a 2 opposite a tg45o = = = 1 adjacent a adjacent a ctg45o = = = 1 opposite a In this way we get table: 30o 45o 60o sinα 1 2 3 2 2 2 cosα 3 2 1 2 2 2 tgα 3 1 3 3 ctgα 3 1 3 3 Of course, we will table later expand to all corners from0o → 360o . Basic trigonometric identity: 1)sin 2 α + cos2 α = 1 sinα 2) tgα = cosα cosα 3) ctgα = sinα 4) tgα ⋅ctgα =1 To try to prove some of the identity: a b a 2 b 2 c2 1) sin2 α + cos 2 α = (by definitions: sinα = and cosα = ) = + = = 1 c c c 2 c 2 c2 a sinα a ⋅c a 2) = c = = = tgα cosα b b ⋅c b c a a a b 4) tgα ⋅ctgα = ( change in the definition, that is tgα = i ctgα = ) = ⋅ =1 b b b a From the basic identity can be done other equality: If you go by: sin 2 α + cos2 α =1 → devide this with cos2 α sin 2 α cos2 α 1 tgα = + = cos2 α cos2 α cos2 α 1 tg 2α +1 = → Express from here cos2 α cos2 α 1 cos2 α = tg 2α +1 Now this change in: sin 2 α + cos2 α =1 1 sin 2 α + =1 tg 2α +1 1 sin 2 α =1− tg 2α +1 tg 2α +1−1 sin 2 α = tg 2α +1 tg 2α sin 2 α = tg 2α +1 These two we will remember and use them in tasks! Next: C b a α β c A B From the image (by definition) is: a b sinα = sin β = c c b a cosα = cos β = c c a b tgα = tgβ = b a b a ctgα = ctgβ = a b We have new identity: sin( 90 o − α ) = cos α sin β = cosα cos(90o −α) = sinα cos β = sinα o tgβ = ctgα tg(90 −α) = ctgα ctgβ = tgα ctg(90o −α) = tgα 1) We have cathetus of right angle triangle: a=8cm and b=6cm. Determine the value of all trigonometric functions for angles α and β. Solution: a = 8 cm a 8 4 sinα = = = = cos β b = 6cm c 10 5 __________ b 6 3 c 2 = a 2 + b 2 cosα = = = = sin β c 10 5 c 2 = 82 + 62 a 8 4 2 tgα = = = = ctgβ c = 64 + 36 b 6 3 c 2 = 100 b 6 3 ctgα = = = = tgβ a 8 4 c = 10cm 2) Calculate the value of trigonometric functions slope angle diagonal of the cube to basis. Solution: As we know , the small diagonal is d = a 2 ,a large (body) diagonal is: D = a 3 . By the definitions: a 1 1 3 3 sin α = = =⋅= a 3 3 33 3 a 2 2 23 6 cos α = = =⋅= a 3 3 33 3 a 1 1 2 2 tg α = = =⋅= a 2 2 22 2 a 2 ctg α = = 2 a 3) Calculate unknown, if: c = 24cm sinα = 8,0 ______________ a = ? b = ? Solution: a sinα = c a 8,0 = 24 a = 24⋅ 8,0 a =19 2, cm b2 = c2 − a2 Pythagorean theorem b2 = 242 − (19 )2, 2 b2 = 576 − 368,64 b2 = 207,36 b = 14 4, cm 4) Calculate the value of other trigonometric functions, if: a) sinα = 6,0 12 b) cosα = 13 c) tgα = ,0 225 Solution: 3 6 3 a) sinα = because 6,0 = = and we will use sin 2 α + cos2 α =1 5 10 5 2 3 + cos2 α =1 5 9 cos2 α =1− 25 16 cos2 α = 25 16 cosα = ± 25 3 4 sinα 3 cosα = ± tgα = = 5 = 5 cosα 4 4 As the sharp angle is in question: 5 4 1 4 cos α = + ctgα = = 5 tgα 3 b) 12 cosα = 13 sin 2 α + cos2 α =1 2 12 sin 2 α + =1 13 144 sin 2 α =1− 169 25 sin 2 α = 169 25 sinα = ± 169 5 sinα 5 5 tgα = = 13 = sinα = ± cosα 12 12 13 13 12 5 ctgα = sin α = 5 13 225 9 c) tgα = ,0 225 = = 1000 40 2 2 tg α sin α = 2 tg α +1 2 9 2 40 sin α = 2 9 +1 40 81 sin 2 α = 1600 81 +1 1600 81 sin 2 α = 1600 81+1600 1600 81 sin 2 α = 1681 81 sinα = ± 1681 9 sinα = ± 41 9 sinα = + 41 1 cos2 α = tg 2α +1 1 cos2 α = 1681 1600 1600 cos2 α = 1681 1600 cosα = ± 1681 40 cosα = ± 41 40 cosα = + 41 1 ctgα = tgα 40 ctgα = 9 5) Calculate the value of other trigonometric functions if: a2 − 9 a) sinα = a2 + 9 a2 − 4 b) ctgα = 4a Solution: a) 36a2 2 2 cos2 α = sin α + cos α = 1 2 2 sin α (a + )9 tg α = cos2 α =1− sin 2 α 2 cos α 36a 2 2 2 cosα = a − 9 2 a − 9 2 2 cos α =1− (a + )9 2 a2 + 9 a + 9 6a tg α = 2 2 cosα = 6a 2 2 (a − )9 a + 9 2 cos α =1− a + 9 (a2 + )9 2 a2 − 9 (a2 + )9 2 − (a2 − )9 2 tg α = cos2 α = 6a (a2 + )9 2 6a 4 2 4 2 ctg α = a +18a + 81− a +18a −81 2 cos2 α = a − 9 (a2 + )9 2 b) a 2 − 4 4a ctgα = ⇒ tgα = 4a a2 − 4 1 cos2 α = tg 2α tg 2α +1 sin 2 α = 2 tg α +1 2 1 cos α = 2 2 4a 4a 2 +1 a 2 − 4 a − 4 sin 2 α = 2 1 4a cos2 α = +1 2 2 2 a2 − 4 16a + (a − )4 2 2 2 (a − )4 16a 2 2 1 2 (a − )4 2 sin α = cos α = 2 2 16a2 (a + )4 +1 2 2 (a2 − )4 2 (a − )4 2 2 2 (a − )4 2 16a 2 sin α = cos α = 2 2 16a2 + a 4 −8a2 +16 (a + )4 2 2 2 2 16a (a − )4 sin α = cosα = a4 + 8a2 +16 (a2 + )4 2 2 16a a2 − 4 sinα = 4 2 cosα = (a + )4 a2 + 4 4a sinα = 2 a + 4 1 1 6) Prove that: 1+ tgx + ⋅1+ tgx − = 2tgx cos x cos x Solution: 1 1 1+ tgx + ⋅1+ tgx − = cos x cos x sin x 1 sin x 1 1+ + ⋅1+ − = cos x cos x cos x cos x cos x + sin x +1 cos x + sin x −1 ⋅ = cos x cos x (cos x + sin x)2 −12 = ( 1 will be replaced with sin 2 x + cos2 x ) cos2 x cos2x+ 2cos xxxx sin +−− sin 222 sin cos x 2cos x sin x = = 2 2 cos x cos x sin x =2 = 2 tgx cos x 7) Prove that : a) cos2 18o + cos2 36o + cos2 54o + cos2 72o = 2 b) tg1o ⋅tg2o ⋅tg3o...tg44o ⋅tg45o ⋅tg46o...tg89o = 1 Proof: a) cos2 18o + cos2 36o + cos2 54o + cos2 72o = 2 Because α + β = 90o , cosα = sin β , cos54o replace with sin36o , and cos72o replace with sin18o . Then: cos2 18o + cos2 36o + cos2 54o + cos2 72o = =1+1 = 2 b) tg1o ⋅tg2o ⋅tg3o...tg44o ⋅tg45o ⋅tg46o...tg89o = = Kako je tgα = ctgβ (α + β = 90o ) Biće= tg1o ⋅tg2o ⋅tg3o...tg44o ⋅tg45o ⋅ctg44o...ctg2o ⋅ctg1o = As is : tgα ⋅ctgα =1 =1⋅1⋅...⋅tg45o = 1 3 8) Prove that : = (tgα + ctgα) 2 1− sin 6 α − cos 6 α Proof: 3 3 = = 1− sin 6 x − cos6 x 1− (sin 6 x + cos6 x) We will attempt to transform expression sin 6 x − cos6 x … sin 2 x − cos2 x =1 (A + B)3 = A3 + 3A2 B + 3AB2 + B3 sin 2 x + cos2 x =1/()3 sin 6 x + 3sin 4 x cos2 x + 3sin 2 x cos4 x + cos6 x =1 sin 6 x + 3sin 2 x cos2 x(sin 2 x + cos2 x) + cos6 x =1 1 So: sin 6 x + cos6 x = 1− 3sin 2 x cos2 x Let's go back to the task: 3 3 3 3 1 = = = = = 1− sin 6 x − cos6 x 1− (sin 6 x + cos6 x) 1−1+ 3sin 2 x cos2 x 3sin 2 x cos2 x sin 2 x cos2 x To see now right side of equation: (tgα + ctgα)2 = tg 2α + 2tgαctgα + ctg 2α sin 2 α cos2 α = + 2 + cos2 α sin 2 α sin 4 α + 2sin 2 α cos2 α + cos4 α = 2 2 sin α cos α (sin 2 α + cos2 α)2 = sin 2 α cos2 α 1 = sin 2 α cos2 α This we have proven that the left and right side are equal.