Mathematics 5 SN

TRIGONOMETRY - IDENTITIES

Peter and Lucy are studying the repeated up and down movement of a piston in a gasoline engine.

Peter claims that the piston moves according to the rule :

t

tt

t

ttd

cos

sincos

sin

cossin2 22

Lucy believes that the piston moves according to the rule :

d = sec t

Prove that Peter and Lucy are both correct by showing that

tt

tt

t

ttsec

cos

sincos

sin

cossin2 22

Note : t = time in seconds and d = distance in metres.

Show your work.

Work

tt

tt

t

ttsec

cos

sincos

sin

cossin2 22

1

Prove that,

xx

x

x

xx 2

2

2

2

22

tancos

cos1

1sec

1tansin

Show your work.

2

Work

x

x

x

x

xx 2

2

2

2

22

tancos

cos1

1sec

1tansin

Prove the following identity :

θsec2θsin1

θcos

θcos

θsin1

.

Show your work.

Work

θsec2θsin1

θcos

θcos

θsin1

3

Prove the following identity :

(1 + cot2)(1 cos2

) = 1.

Show your work.

Work

(1 + cot2)(1 cos2

) = 1

4

Prove the following identity :

sec2 cot2

1 = cot2.

Show your work.

Work

sec2 cot2

1 = cot2

5

Prove the following identity :

θsinθtan

θcosθsec

.

Show your work.

Work

θsinθtan

θcosθsec

6

Prove the following identity :

tan2 sin2

= sin2 tan2

.

Show your work.

Work

tan2 sin2

= sin2 tan2

.

7

Prove the following identity :

θcosecθsinθtan

θsec1

.

Show your work.

Work

θcosecθsinθtan

θsec1

8

Prove the following identity :

θcosθtan

θcos1 2

2

2

.

Show your work.

Work

θcosθtan

θcos1 2

2

2

9

Prove the following identity :

sec cos (sec 1) = cos .

Show your work.

Work

sec cos (sec 1) = cos .

10

Which expression is equivalent to (sec2t 1)(cosec2t 1)?

A)

0

C)

sin t

B)

1

D)

cos t

Which expression is equivalent to (1 sin2t)(1 cos2t)?

A)

0

C)

2sin2t cos2t

B)

1

D)

sin2t cos2t

If tan = ,3

5 what is the value of the following functions :

a) sin

b) cos

c) sec

d) cosec

e) cot

11

12

13

a) _________________________

b) _________________________

c) _________________________

d) _________________________

e) _________________________

Which expression is equivalent to sin2t sec2t sec2t?

A)

-cot2t

C)

0

B)

-1

D)

1

Which expression is equivalent to(sin x cos x)2?

A)

1

C)

sin2x cos2x

B)

1 2sin x cos x

D)

2sin2x

14

15

2- Correction key

Work : (example)

1.

t

tt

t

tt

cos

sincos

sin

cossin2 22 = sec t

2.

t

ttt

cos

sincoscos2

22 =

3.

t

ttt

cos

sincoscos2 222 =

4.

t

tt

cos

sincos 22 =

5.

tcos

1 =

6.

sec t =

1

Work : (example)

1.

1sec

1tansin2

22

x

xx

x

x2

2

cos

cos1 = tan2x

2. x

xx2

22

tan

secsin

x

x2

2

cos

sin = tan2x

3. x

xx2

22

tan

secsin tan2x = tan2x

4. sin2x sec2x = tan2x

5. sin2x x2cos

1 = tan2x

6. tan2x = tan2x

2

Work : (example)

θsin1

θcos

θcos

θsin1

= 2 sec

θsin1θcos

θcosθsin1 22

=

2 sec

θsin1θcos

θcosθsinθsin21 22

=

2 sec

θsin1θcos

1θsin21

=

2 sec (because sin2 + cos2

= 1)

θsin1θcos

θsin22

=

2 sec

θsin1θcos

θsin12

=

2 sec

θcos

2 =

2 sec

θsec

1θcosbecause

2 sec =

2 sec

3

Work : (example)

(1 + cot2)(1 cos2

) = 1

cosec2(1 cos2

) = 1

cosec2(sin2

) = 1

1θsinθsin

1 2

2

1 = 1

as 1 + cot2 = cosec2

as cos2 + sin2

= 1

as θcosec

1θsin

4

Work : (example)

sec2 cot2

1 = cot2

θcot1θcotθcos

1 22

2

θcot1θsin

θcos

θcos

1 2

2

2

2

θcot1θsin

1 2

2

cosec2 cot2

cot2 = cot2

as θsec

1θcos

as θsin

θcosθcot

as θcosec

1θsin

5

Work : (example)

θsinθtan

θcosθsec

θsinθtan

θcosθcos/1

θsinθtanθcos

θcos1 2

θsinθsinθcos

θcosθsin 2

θsinθsin

as θsec

1θcos

as sin2 + cos2

= 1

6

Work :

Example 1

tan2 sin2

= sin2 tan2

sin2(sec2

1) = sin2 tan2

sin2(tan2

+ 1 1) = sin2 tan2

sin2 tan2

= sin2 tan2

Example 2

tan2 sin2

= sin2 tan2

θtanθsinθsinθcos

θsin 222

2

2

θtanθsinθcos

θsinθcosθsin 22

2

222

θtanθsin

θcos

θsinθcos1 22

2

22

sin2 tan2

= sin2 tan2

7

Work : (example)

θcosecθsinθtan

θsec1

θcosecθsinθcos/θsin

θsec1

θcosec

θsinθsecθsin

θsec1

θcosec

θsec1θsin

θsec1

θcosecθsin

1

θcosecθcosec

as θsec

1θcos

as θcosec

1θsin

8

Work : (example)

θcosθtan

θcos1 2

2

2

θcosθtan

θsin 2

2

2

θcosθcos/θsin

θsin 2

22

2

cos2 = cos2

as sin2 + cos2

= 1

as θcos

θsinθtan

9

Work : (example)

sec cos (sec2 1) = cos

sec cos tan2 = cos

θcosθcos

θsinθcosθsec

2

2

θcosθcos

θsinθsec

2

θcosθcos

θsin

θcos

1 2

θcosθcos

θsin1 2

θcosθcos

θcos 2

cos = cos

as tan2 + 1 = sec2

as θcos

θsinθtan

as θcos

1θsec

as sin2 + cos2

= 1

10

B

D

a) 34

5 or 0.8575 b)

34

3 or 0.5145 c)

3

34 or 1.9437

d) 5

34 or 1.1662 e)

5

3 or 0.6

B

B

11

12

13

14

15