5.2
Verifying Trigonometric
Identities
2
sec x ! 1
2
= sin x
2
sec x
2
sec x
1
!
=
2
2
sec x sec x
1 - cos2 x =
sin2 x = sin2 x
Place sec2 x under both
terms.
1
1 - sin x
+
1
1 + sin x
=
2 sec2 x
1 + sin x + 1 ! sin x
=
(1 ! sin x )(1 + sin x )
2
=
2
1 ! sin x
2
=
2
cos x
2
2
2 sec x = 2 sec x
(tan2 x + 1)(cos2 x - 1) = -tan2 x
(sec2 x)(-sin2 x) =
2
sin x
!
=
2
cos x
2
2
! tan x = ! tan x
tan x + cot x = sec x csc x
sin x cos x
+
=
cos x sin x
Get common denominators.
sin 2 x + cos 2 x
=
cos x sin x
1
=
cos x sin x
sec x cscx =
sec x csc x
cos y
sec y + tan y =
1 ! sin y
1
sin y
+
=
cos y cos y
1+ sin y
cos y
& 1 ' sin y #
$$
!! =
% 1 ' sin y "
1 ! sin 2 y
=
cos y (1 ! sin y )
cos y cos y cos y
=
cos y (1 ! sin y ) 1 ! sin y
2
We need a (1 – sin y) in the
denominator so let’s put
one there.
cot 2 x 1 ! sin x
=
1 + csc x
sin x
2
csc x ! 1
=
1 + csc x
(csc x ! 1)(csc x + 1)
=
1 + csc x
csc x – 1 =
1
1!
sin
x
!1 =
sin x
sin x
tan4 x = tan2 x sec2 x - tan2 x
(tan2 x)(tan2 x) =
Look at the right side.
How can we break up
tan4 x?
tan2 x(sec2 x - 1) =
tan2 x sec2 x - tan2 x = tan2 x sec2 x - tan2 x