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Trigonometry was originally an area of mathematics that deal with calculating unknown elements of the triangle

with wellknown. The name comes from two Greek words , TRIGONOSmeaning triangle and METRONmeaning measures. How do we define trigonometric functions?

Observe rightangled triangle ABC.

a, b → cathetus c → a2 + b2 = c2 →

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 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