1

Lesson 6

Trigonometric Function (III)　

6A •  Trigonometric Identities •  Exact Values of Some Trigonometric Function

Course I

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2

Sum and Difference Identities for Sine and Cosine

Sum and Difference Formulae

βαβαβαβαβαβαβαβαβαβαβαβα

sinsincoscos)cos(sinsincoscos)cos(sincoscossin)sin(sincoscossin)sin(

+=−−=+−=−+=+

An identity is an equality that is true for any value of the variable. (An equation is an equality that is true only for certain values of the variable.)

In calculation, we may replace either member of the identity with the other. We use an identity to give an expression a more convenient form.

This formula is most important because all the identities in the following are based on this formula.

Memorize !

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Proof of Sum Formula for Sine

)sin( βα +DE = AD =FG + DH =FA +DF =AD + AD

αsinαcosαβ sincos βsin

αcos

βαβαβα sincoscossin)sin( +=+

Therefore

Example 1 Prove the formula βαβαβα sincoscossin)sin( +=+

We can prove these identities geometrically.

From the figure

Ans.

4

Find Exact Values of Trigonometric Function We can find values of some trigonometric functions from known cases.

2 1

1

12

3

Example 2 Find the values of (a) sin15°and (b) cos15°. Ans.

426

2213

21

21

23

21

sin30sin45cos30cos45)3045cos(15cos

,4

262213

21

21

23

21

30sin45cos30cos45sin)3045sin(15sin

+=

+=×+×=

°°+°°=°−°=°

−=

−=×−×=

°°−°°=°−°=°

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Sum and Difference Identities for Tangent Example 3 Derive the sum and difference formulae of the tangent function. 　　　　　　　

βαβα

βαβαββ

αα

βαβαβαβα

βαβα

βαtantan1tantan

coscossinsin1

cossin

cossin

sinsincoscossincoscossin

)cos()sin()tan(

−

+=

−

+=

−

+=

+

+=+

βαβα

βαβαββ

αα

βαβαβαβα

βαβα

βαtantan1tantan

coscossinsin1

cossin

cossin

sinsincoscossincoscossin

)cos()sin()tan(

+

−=

+

−=

+

−=

−

−=−

βα coscos÷

In summary

βαβα

βαβαβα

βαtantan1tantan)tan(,

tantan1tantan)tan(

+

−=−

−

+=+

We use the sine and cosine formulae.

6

Exercise

Pause the video and solve the problem.

Ans. Exercise 1 　Find the values of (a) sin75°, (b) cos75°and (c) tan75°.

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Answer to the Exercise

Ans. Exercise 1 　Find the values of (a) sin75°, (b) cos75°and (c) tan75°.

426

2213

21

21

23

21

30sin45cos30cos45sin)3045sin(75sin

+=

+=×+×=

°°+°°=°+°=°

426

2213

21

21

23

21

30sin45sin30cos45cos )3045cos(75cos−

=−

=×−×=

°°−°°=°+°=°

1313

3/1113/11

30tan45tan130tan45tan )3045tan(75tan

−

+=

×−

+=

°°−°+°

=°+°=°

(a) (b)

(c)

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Exercise

Ans.

Exercise 2 There are two lines and Answer the following questions. (1)  Let the angles made by these lines and the x-axis by α and β. Express the slopes of these lines by tangent function values. (2)  Find the angle θ (0<θ< ) made by these two lines.

12113 +=−= xyxy

2π

Pause the video and solve the problem.

Uhh….

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Exercise

Ans.

Exercise 2 There are two lines and Answer the following questions. (1)  Let the angles made by these lines and the x-axis by α and β. Express the slopes of these lines by tangent function values. (2)  Find the angle θ (0<θ< ) made by these two lines.

12113 +=−= xyxy

2π

21tan,3tan)1( == βα

βαθ −=)2(

4,1

2131213

tantan1tantan)tan(tan π

θβαβα

βαθ =∴=×+

−=

+

−=−=

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Lesson 6

Trigonometric Function (II)　

６B6B

•  Derivation of Trigonometric Identities •  Double-Angle Identities •  Half-Angle Identities •  Product-to-Some Identities •  Some-to-Product Identities

Course I

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Derivation of Trigonometric Identities

βα =

Double-Angle Identities etc.

Half-Angle Identities etc.

αβ 2=

Three-Angle Identities etc.

Product-to-Sum Identities etc.

Sum-to-Product Identities etc.

Solve inversely Put

2α

α →

βαβαβαβαβαβαsinsincoscos)cos(

sincoscossin)sin(∓=±

±=±

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Double-Angle Identities

Example 4 Prove the double angle identity of tangent

Ans.

αα

αααα

αααα

αα

α

2

22

tan1tan2

sincoscossin2

)cos()sin(

2cos2sin2tan

−=

−=

+

+=

= : From definition

: From sum identities

: Divide by and use

α2cos

αα

αcossintan =

1cos2sin21sincoscos2

cossin2sin2

22

22

−=−=

−=

=

αα

ααα

ααα

αα

α 2tan1tan22tan

−=

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Half-Angle Identities

Example 5 Evaluate the values of

Ans. Since 22.5°is in quadrant I, all these values are positive.

Therefore

2225.22cos

422

245cos15.22cos

2225.22sin

422

22122/

211

245cos15.22sin

2

2

+=∴

+=

+=

−=∴

−=

−=⎟

⎠

⎞⎜⎝

⎛−=

−=

!!

!

!!

!

12222

24)22(

22

225.22cos5.22sin5.22tan

2

−=−

=−−

=+

−== !

!!

2cos1

2cos 2 αα +

=ααα

cos1cos1

2tan2

+

−=2

cos12

sin 2 αα −=

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Product-to-Sum Identities

Example 6 Evaluate the values of

Ans.

)}cos(){cos(21sinsin

)}cos(){cos(21coscos

)}sin(){sin(21sincos

)}sin(){sin(21cossin

βαβαβα

βαβαβα

βαβαβα

βαβαβα

−−+−=

−++=

−−+=

−++=

41cos60

21)}60cos({cos90

21

)}7515cos()7515{cos(2175sin15sin

=°=°−−°−=

°−°−°+°−=°°

°° 75sin15sin

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Sum-to-Product Identities

Example 7 Evaluate the values of Ans.

P-to-S

S-to-P

BA

=−=+

βαβα

2

sin2

sin2coscos

2cos

2cos2coscos

2sin

2cos2sinsin

2cos

2sin2sinsin

BABABA

BABABA

BABABA

BABABA

−+−=−

−+=+

−+=−

−+=+

°+°+° cos230cos110cos10

0cos110cos110 cos110)110cos(2cos120

cos110)cos230(cos10cos230cos110cos10

=°+°−=°+°−°=

°+°+°=°+°+°

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Exercise

Ans.

Pause the video and solve the problem.

Exercise 3　Let　 and　　　　　　　　．　Express the following functions by　 and 　 . (1) (2)　

θθ cossin == cscs

θθ 4cos4sin

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Answer to the Exercise Exercise 3　Let　 and　　　　　　　　．　Express the following functions by　 and 　 . (1) (2)　

θθ cossin == cscs

θθ 4cos4sin

Ans.

18c8c11)4c2(4c 1)12(2

11)2(2cos122cos)2cos(2cos4 (2)

)2ssc(14cos2 2sin2)2sin(2sin4 )1(

2424

22

222

2

+−=−+−=

−−=

−−=−=×=

−==×=

cθθθθ

θθθθ

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Exercise

Ans.

1cossin 22 =+ xxExercise 4 When , evaluate [Hint] Divide the expression by .

2tan −=x xx 2cos2sin +

Pause the video and solve the problem by yourself.

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Answer to the Exercise

Ans.

1cossin 22 =+ xxExercise 4 When , evaluate [Hint] Divide the expression by .

2tan −=x xx 2cos2sin +

57

12)(2)(12)2(

1tantan12tan

cossinsincoscos2sin

12cos2sin2cos2sin

2

2

2

2

22

22

−=+−−−+−

=+−+

=

+−+

=+

=+

xxx

xxxxxxxxxx