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 2 2 2 c = 8 + 6 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
tg 2α 2 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 = tg 2α +1 1 cos2 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 16a2 2 2 sin 2 α = (a − )4 4 2 cosα = 2 2 a + 8a +16 (a + )4 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.
1− sin6 α − cos6 α ≠ 0 sin 6 α − cos6 α ≠ 1 Condition is: 1− 3sin 2 α cos2 α ≠ 1 sin 2 α cos2 α ≠ 0 sinα ≠ 0 ∧ cosα ≠ 0
1−tgα 1 − ctg α 9) Prove that : (tg3α+ ):( + ctg3 α ) = tg 4 α ctgα tg α
Proof:
1−tgα 1 − ctg α (tg3α+ ):( + ctg 3 α ) = ctgα tg α 1 1− 1− tg α tg α 1 (tg 3α + ):( + ) = 1 tgα tg 3 α tg α tg α −1 tg α 1 (tg3α+⋅− tg α (1 tg α )) : ( + ) = tgα tg 3 α tg α −1 1 (tg3α+− tg α tg 2 α ):( += ) tg2α tg 3 α tgα( tg α − 1) + 1 (tg3α− tg 2 α + tg α ):( ) = tg 3α tg 2α − tgα+1 tgtg ααα (2 −+ tg 1) tg 2 αα −+ tg 1 tgα( tg2 α− tg α + 1) : ( )= :( ) = tg3α1 tg 3 α tgα( tg2 α− tg α + 1) tg 3α ⋅ =tgα ⋅ tg3 α = tg 4 α 1 tg2α− tg α + 1
Conditions are tg α ≠ 0 and ctg α ≠ 0