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Example
Solution
Example
Solution
Example
Solution
Example
Solution
Example
Solution
Example
4
If cos 𝑥 = − and tan x is positive, find the value of sin x.
5
Solution
Since cos x is negative and tan x is positive, x is in the third quadrant. In
right ΔPOQ, the length of ̅̅̅̅
𝑂𝑄 is 4 and of ̅̅̅̅
𝑂𝑃 is 5. It follows that the
3
length of PO is 3. Therefore, sin x = −
5
Example
Solution
Example Express sin2x and cos2x in terms of tan x.
Solution
2
sin x =
cos2x =
sin2 𝑥
1
cos2 𝑥
1
=
=
sin2 𝑥
sin2 𝑥 + cos2 𝑥
cos2 𝑥
sin2 𝑥 + cos2 𝑥
=
=
Thus,
sin2x =
cos2x =
tan2 𝑥
tan2 𝑥 + 1
1
tan2 𝑥 +
1
sin2 𝑥
cos2 𝑥
2
sin 𝑥
cos2 𝑥
+
cos2 𝑥
cos2 𝑥
cos2 𝑥
cos2 𝑥
2
sin 𝑥
cos2 𝑥
+
2
cos 𝑥
cos2 𝑥
=
=
tan2 𝑥
,
tan2 𝑥 + 1
1
tan2 𝑥 +
.
1
Example What is the value of sin (arctan √2 )?
Solution Let 𝜃 = arctan √2, then 0° < 𝜃 < 90° and tan 𝜃 = √2.
Hence,
2
sin 𝜃 =
tan2 𝜃
tan2 𝜃
+ 1
2
=
(√2)
2
(√2) + 1
=
2
3
2
⇒ sin 𝜃 = √ =
3
√6
.
3
Example Find the value of sin (arcsin 1 + arccos 1).
Solution arcsin 1= 90°, arccos 1= 0°, therefore,
sin (arcsin 1 + arccos 1) = sin (90° + 0°) = sin 90° = 1.
Example Prove the identity:
1 + sec x – cos x – sec x cos x = tan x sin x
Solution
1 + sec x – cos x – sec x cos x = (1 + sec x) – cos x (1 + sec x)
= (1 + sec x) (1– cos x)
= (1 +
= (1 +
=
=
=
1
cos 𝑥
1
cos 𝑥
) (1– cos x)
) (1– cos x)
1− cos2 𝑥
cos 𝑥
sin2 𝑥
cos 𝑥
sin 𝑥
cos 𝑥
∙ sin 𝑥
= tan 𝑥 ∙ sin 𝑥
Example If p = cos 200°, express cot 70° in terms of p.
Solution
cos 200° = cos (270° − 70°) = −sin 70° ⇒ sin 70° = −𝑝,
cos 70° = √1 − sin2 70° = √1 − (−𝑝)2 = √1 − 𝑝2 ,
cot 70° =
cos 70°
sin 70°
= −
√1− 𝑝2
𝑝
.
sec 𝑥
Example The expression
is equivalent to
tan 𝑥 sin 𝑥
(A) tan x
(B) csc 2 𝑥
(C) sec 2 𝑥
(D) sec 2 𝑥 csc 𝑥
(E) cos x
Solution Answer: (B)
sec 𝑥
=
tan 𝑥 sin 𝑥
1
cos 𝑥
sin 𝑥
∙ sin 𝑥
cos 𝑥
=
1
cos 𝑥
sin2 𝑥
cos 𝑥
1
= cos 𝑥 ∙
cos 𝑥
sin2 𝑥
=
1
sin2 𝑥
= csc2 𝑥.
Example
Solution
Example
Solution
Example
sin 2𝜃
Express
2 as a trigonometric function of 𝜃.
2sin 𝜃
Solution
sin 2𝜃
2 =
2 sin 𝜃
2 sin 𝜃 cos 𝜃
2 sin2 𝜃
=
cos 𝜃
sin 𝜃
= cot 𝜃.
Example
3
If sin x = and x is an acute angle, find the value of sin 2x.
5
Solution
3
sin x = , 0° < 𝑥 < 90° ⇒
5
cos x = √1 −
sin2 𝑥
2
3
= √1 − ( ) = √1 −
5
3
4
5
5
sin 2x = 2 sin x cos x = 2 ∙ ∙
=
24
25
9
4
25
= ,
5
.
Example
1
𝑥
If cos x = , 360° < 𝑥 < 450°, what is the value of sin ( ) ?
9
2
Solution
𝑥
𝑥
360° < 𝑥 < 450° ⇒ 180° < < 225° ⇒ sin ( ) < 0,
2
sin
2 𝑥
( )
2
=
1 − cos 𝑥
2
=
1−
2
1
9
=
2
4
9
𝑥
2
2
3
⇒ sin ( ) = − .
Example Prove the formulas:
𝑥
1 − cos 𝑥
2
sin 𝑥
𝑥
sin 𝑥
2
1 + cos 𝑥
tan ( ) =
tan ( ) =
Solution
1 − cos 𝑥
sin 𝑥
sin 𝑥
1 + cos 𝑥
𝑥
=
𝑥
𝑥
2 sin(2 ) cos(2)
𝑥
=
𝑥
2sin 2 ( )
2
𝑥
2 sin(2 ) cos(2)
𝑥
2cos 2 (2)
=
sin( )
2
𝑥
cos(2)
𝑥
=
sin(2)
𝑥
cos(2)
𝑥
= tan ( ),
2
𝑥
= tan ( ).
2
Example Prove the formulas:
𝑥
sin 𝑥 =
2 tan(2)
𝑥
1 + tan2 (2)
𝑥
cos x =
1− tan2 (2)
𝑥
1 + tan2 (2)
Solution
𝑥
𝑥
2 sin( ) cos( )
2
2
2
2
sin2 (2) + cos2 (2 )
sin 𝑥 = 2 sin ( ) cos ( ) =
=
𝑥
𝑥
2 sin( ) cos( )
2
2
𝑥
2
cos ( )
2
𝑥
𝑥
sin2 ( )+ cos2 ( )
2
2
𝑥
cos2 ( )
2
cos x = cos
=
𝑥
𝑥
2 𝑥
( )
2
𝑥
𝑥
cos 2 ( ) − sin 2 ( )
2
2
𝑥
cos2 ( )
2
𝑥
𝑥
2
sin ( )+ cos2 ( )
2
2
𝑥
cos2 ( )
2
𝑥
𝑥
=
− sin
2 tan( )
2
𝑥
,
1 + tan2 ( )
2
2 𝑥
( )
2
=
𝑥
=
𝑥
1− tan2 (2)
𝑥
,
1 + tan2 (2)
𝑥
𝑥
𝑥
𝑥
cos 2 (2) − sin 2 (2)
sin2 (2) + cos2 (2)
Example
4
8
5
17
If sin x = , 90° < 𝑥 < 180°, and cos y =
, 0° < 𝑦 < 90°, what is the
value of cos (x + y)?
Solution
4
sin x = , 90° < 𝑥 < 180° ⇒
5
cos x = −√1 −
cos y =
8
sin2 𝑥
2
4
3
= −√1 − ( ) =− ,
5
5
, 0° < 𝑦 < 90° ⇒
17
8 2
15
17
17
sin y = √1 − cos 2 𝑦 = √1 − ( ) =
cos (x + y) = cos x cos y – sin x sin y
3
8
4
15
5
17
5
17
= (− ) ( ) − ( ) ( )
=−
24
85
−
60
85
= −
84
.
85
,