C3 Trigonometry 2o1i6n

C3 Trigonometry

-1-

Secant, Cosecant, Cotangent 1.

Sec x =

1 , cos x

cos x ≠ 0

2.

Cosec x =

1 , sin x

sin x ≠ 0

3.

Cot x =

1 , tan x

tan x ≠ 0

=

cos x , sin x

sin x ≠ 0

On your calculator, you will have to use sin, cos and tan keys, followed by the 1 −1  reciprocal key  or x  . However, to find exact values for the 'special' x  angles, follow the following examples. Examples Find the exact value of: 1

(i) (ii) (iii) (iv)

2.

(i) (ii) (iii)

sec 60 o cosec 60 o cot 30 o cosec 45 o π cosec 6 π sec 4 π cot 3

Solution With calculaor in degrees 1. (i) (ii)

1   Press cos 60  because sec = to get 0.5 then press 1/x to get 2 cos   press sin 60 to get 018660 ... which isn't a recognisable fraction ... then press x2 to get 0.75 ( = 34 ) i.e. sin2 60 =

3 4

/opt/idoub/conversion/tmp/scratch25351/31244900.doc 1

C3 Trigonometry

-2-

3 2

so, sin 60 =

2

so, cosec 60 = (iii)

3

press tan 30 to get 05773 ... which isn’t a recognisable fraction ... then press to get 0.3333 ... which is 31 . x2 i.e. tan2 30 = so, tan 30 =

1 3 1 3

so, cot 30 o =

(iv)

3

press sin 45 o to get 0.7071 ... which is not a recognisable fraction ... then press x2 to get 0.5 ( = 21 ) i.e. sin2 45 =

1 2 1

so, sin 45 =

2

so, cosec 45 =

2

with calculator in rads: 2. (i)

press sin ( π ÷ 6) to get 0.5 ( =

1 2

)

∴ cosec

π =2 6

(ii) press cos ( π ÷ 4) to get 0.7071 ... then press i.e.

cos2

so, cos

sec,

π = 4

π = 4

π = 4

x2

1 2

1 2

2


to get 0.5 ( =

1 2

).

C3 Trigonometry (iii)

-3-

Press tan ( π ÷ 3 ) to get 1.732 ... then press i.e. tan2

x2

π =3 3

tan

π = 3

cot

1 π = 3 3

3

The graphs of sec, cosec and cot are shown below:

y = cosecθ

θ

y = secθ θ


to get 3

C3 Trigonometry

-4-

y = cot θ

Their period are the same as cos, sin and tan. Also,

Sin+

All+

Tan+

Cos+

⇒

Cosec+

All+

Cot+

Sec+

Trigonometric Identities You already know sin2 θ + Cos2 θ ≡ 1 If you divide by Cos θ 2

2 1 → sin 2θ +1 ≡ cos2 θ cos θ

We get

tan2 θ + 1 ≡ sec2 θ

similarly, dividing by sin2 θ gives

1 + cot2 θ ≡ cos ec 2θ

These can be used to simplify expressions


C3 Trigonometry

-5-

Example (i)

Express 5cot θ + 2 cosec2 θ in of cot θ

(ii)

Solve the equation 5 cot θ + 2 cosec2 θ = 5 for 0 < θ < 2 π

Solution (i)

5 cot θ + 2 cosec2 ≡ 5 cot θ + 2 (1 + cot2 θ ) ≡ 5 cot θ + 2 + 2cot2 θ

(ii)

5 cot θ + 2 cosec2 θ = 5 ⇒ 5 cot θ + 2 + 2 cot2 θ = 5 ⇒ 2 cot2 θ + 5 cot θ - 3 = 0

This is a quadratic in cot θ cot θ =

=

−5 ± 25 − 4(2)( −3) 2x2 −5 ± 49 −5 ± 7 1 = = 2 or -3 4 4

⇒ tan θ = 2 or - 31

+ve tan ∴ 1st and 3rd quad.

-ve tan ∴ 2nd and 4th quad

Because the range of θ is given in of π we must have our Calculator in radians

θ = tan −1 2 = 1.11

θ = 1.11, π + 1.11 =1.11, 4.25


C3 Trigonometry

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 1 θ = tan −1  −  = −0.322  3

The Addition Formulae For all angles, A and B, sin ( A ± B ) = sin A cos B mcos A sin B

cos ( A ± B ) = cos A cos B msin A sin B tan ( A ± B ) =

tan A ± tan B 1 mtan A tan B

(these are in the formula book) o o Example 1: Prove the identity sin ( x + 45 ) + cos ( x + 45 ) = 2 cos x

Solution:

sin ( x + 45o ) = sin x cos 45 + cos x sin 45 1 1 + cos x x 2 2 cos ( x + 45) = cos x cos 45 − sin x sin 45 = sin x x

1 1 − sin x x 2 2

= cos x x

cos x cos x + 2 2 1   1 = cos x  +  2  2 2 = cos x x 2 = 2 cos x

∴ sin ( x + 45 ) + cos ( x + 45) =

Example 2: By writing 75o as (30o + 45o) find the exact value of cos75o


C3 Trigonometry

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Solution:

cos75 = cos(30 + 45) = cos30cos45 - sin30sin45 3 2 1 2 = x − x 2 2 2 2 6 2 = − 4 4 Example 3: Find the value of tanxo given that sin(x + 30o) = 2cos(x – 30o) Solution:

sin(x + 30) = 2cos(x - 30) sin x cos 30 + cos x sin 30 = 2[cos x cos 30 + sin x sin 30] 3 1 3 1 sin x + cos x = 2 [ cos x + sin x] 2 2 2 2 = 3 cos x + sin x

Rearranging gives 3 1 sin x − sin x = 3 cos x − cos x 2 2  3  1  sin x  = 1 = cos x  3 −  2   2   3   1 tan x  − 1 =  3 −  2  2   1 3− 2 tan x = 3 −1 2 2 3 −1 = 3−2

( ÷ cos x )

 2 ×   2

Double Angle Formulae(LEARN) Putting A=B into the addition formulae gives identities for sin2A, cos2A and tan2A:Using,

sin(A+B) = sin A cos B + cos A sin B.

Sin 2A = sin(A+A) = sin A cos A + cos A sin Sin 2A = 2 sin A cos A Similarly


C3 Trigonometry

-8-

cos 2 A = cos2 A − sin2 A = 2cos2 A − 1 = 1 − 2 sin2 A Note: there are 3 forms of cos2A and tan 2A =

2 tan A 1 − tan2 A

Example: Solve the equation 2 sin 2θ = sin θ for 0 ≤ θ ≤ 360o (to 1dp) Solution: 2 sin 2θ

= sinθ

4 sinθ cos θ

= sinθ

2 [ 2 sinθ cos θ ] = sinθ ( Do not divide by sin θ. You must factorise)

4 sin θ cos θ − sin θ = 0 sin θ (4 cos θ − 1) = 0

sinθ = 0

or

θ = 0,180o,360o

4 cos θ − 1 = 0 cos θ = 0.25

θ = 75.5o,284.5o

θ = 0o,75.5o,180o,284.5o,360o

Example;

If sin A =

3 and A is obtuse, find the exact values of cos A, 5


C3 Trigonometry

-9-

sin2 A, tan 2 A. Solution: sin2 A + cos2 A = 1 ⇒ cos2 A = 1 − sin2 A 9 16 = 1− = 25 25 4 ∴ cos A = ± , but A is obtuse. 5 4 ∴ cos A = − 5 sin2 A = 2 sin A cos A 3 4 = 2 × × (− ) 5 5 24 =− 25 sin A 3 2 tan A tan A = = − ⇒ tan2 A = cos A 4 1 − tan2 A 2( − 34 ) − 32 24 = = 7 =− 9 1 − 16 7 16

Solve cos 2A + 3 + 4cos A = 0, 0 ≤ A ≤ 2π

Example: Solution:

(Since we have cos A in the equation we choose the form of cos 2A which involves cos A only) cos 2 A = 2cos2 A − 1

i.e.

i

Sub in equation : (2cos2 A − 1) + 3 + 4cos A = 0 2cos2 A + 4 cos A + 2 = 0 cos2 A + 2cos A + 1 = 0 (this is a quadratic in cosA) −2 ± 4 − 4 2 = −1

cos A =

A = cos -1 ( −1) A=π The form a cos x + b sin x ≡ R cos( x ± ∝), R sin( x ± ∝ )


C3 Trigonometry

- 10 -

It is useful to be able to express the sum of sin x and cos x as a single sin or cos. It enables us to: - find the max/min of the expression - solve equations To find R and ∝ , we expand the right hand side and equate like . This gives R = + a 2 + b2 ← ( ALWAYS ) ∝ ∝ and values for sin and cos in of a, b and R. Example: i) Express 3sinx+2cos x in the form Rsin( x + ∝) ii) Find the max and min of the expression and find in radians to 2dp the smallest positive value of x for which they occur. iii) Solve the equation 3 sin x + 2 cos x = 1 Solution: (i)

for -π ≤ x ≤ π to 2dp

R= 32 + 22 = 13

3 sin x + 2cos x ≡ R sin( x + ∝) 3 sin x + 2cos x ≡ R sin x cos ∝ +R cos x sin ∝ 3 sin x + 2cos x ≡ (R cos ∝)sin x + (R sin ∝)cos x Equating sin x :→ Equating cos x :→

3 = R cos ∝ ....(1) 2 = R sin ∝ ....(2)

(2) ÷ (1) gives tan ∝=

2 3

2 ∝= tan−1( ) = 0.5880 and R= 13, from above 3 ∴ 3 sin x + 2cos x = 13 sin( x + 0.5880) ii) Max and min of 3 sin x + 2 cos x is same as max & min of 13sin(x+0.5880) Max of any sine is +1 Min of any sine is -1 ∴ Max of expression is + 13 and Min of expression is - 13 Max when sin (x +0.5880) = 1 ∴ x + 0.5880 = π , x = 0.9828 2 = 0.98c (2dp) Min when sin ( x + 0.5880) = −1 ∴ x + 0.5880 = 3π x = 4.12c (2dp ) 2 /opt/idoub/conversion/tmp/scratch25351/31244900.doc 10

C3 Trigonometry

iii )

- 11 -

3 sin x + 2cos x = 1 13 sin( x + 0.5880) = 1 ⊥ 13 1 x + 0.5880 = sin−1( ) 3 x + 0.5880 = 0.2810. sin( x + 0.5880) =

x + 0.5880 = 0.2810, = 0.2810, x = −0.307, = −0.31, NB:

i.e

π − 0.2810 2.8606 2.7226 2.72 (2dp )

3sin x + 2cos x can also be expressed in the form R cos ( x + ∝ ), but it is usually more convenient to choose the form which produces the in the right order with the correct sign. a sin x - b cos x = R sin ( x − ∝) a cos x + b sin x = R cos ( x − ∝) a cos x - b sin x = R cos ( x + ∝)

Inverse Trig Functions For f -1(x) to exist then f ( x ) must be a one to one function. If f(x) Is not 1-to-1 over its whole domain then f −1( x ) can exist provided we restrict the domain of f (x) The functions cos x,sin x and tan x are not 1-to-1 over their whole domain. The inverse functions cos-1 x, sin-1 x and tan−1 x exist for the domains:


C3 Trigonometry

- 12 0≤ x ≤π −π ≤ x ≤ π 2 2 −π π <x< 124 2 4 32

cos−1 x : sin−1 x : tan−1 x :

(tan x has asymptotes at ±

π ) 2

The graph of f −1 (x) is a reflection of f(x) in the line y = x ∴ sin−1,cos−1 x,tan−1 x are reflections of sin x, cos x tan x in the line y = x (x in radians, same scale on both axes).

π 2

π 4

−

π 2

-1

f ( x ) = sin −1 x

1 −

π 4

−

π 2

π 2

f ( x ) = arcsin x π

π 2

−

π 2

-1

1

π 2

f ( x ) = cos −1 x

f ( x ) = arccos x


C3 Trigonometry

- 13 -

π 2

−

π 2

-1

1

−

π 2

π 2

f ( x ) = tan −1 x

f ( x ) = arctan x


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