Math104
NO CALCULATOR
Identity Problems
3
3
and x is in quadrant III and siny =
with y in quadrant I , use the
5
4
identities to find the exact values:
(Be Careful: the second does not make a special triange)
1. If cosx =
a) sin(x + y) = sinxcosy + sinycosx = (
3 3
4
7
4 7 9

)(
) + ( )(
)=
4
5
5
4
20
20
If you draw the triangles corresponding to the angles x and y you can fint the values
above.
b)cos2x = cos2x - sin2x = (-3/5)2 - (-4/5)2 =
c) sin2x = 2sinxcosx = 2 (
7
25
24
4 3
)(
)=
4
5
5
2. Find the exact value of a) cos (165 )
b) tan (15 )
c) sin(22.5⁰) Omit
a) cos(45⁰ + 120⁰)= cos(45⁰) cos(120⁰) - sin(45⁰)sin(120⁰) =
2 1 2
3  2 6
* *
=
2
2
2
2
4
c) Half-angle identity not on this exam.
3. Find sin(67.5˚) using the half-angle identity. Omit - half-angle identity not on this
exam.
3
3
) + cos(x + ) = sinx - cosx
2
2
3
3
3
3
L = sin(x )cos( ) + sin( ) cosx + cos(x)cos( ) - sin(x)sin( )
2
2
2
2
= sin(x )(0) + (-1)cosx + cos(x)(0)- sin(x)(-1) = 0 - cosx + 0 +sinx = R
4. Verify the identity: sin(x +
5. Verify the identity:
L=
=
1
1  sin x
*
1  sin x 1  sin x
1
1  sin x


1
1  sin x
1
1  sin x
*
1  sin x 1  sin x
  2 sec x tan x

1  sin x
1  sin x

2
1  sin x 1  sin 2 x
1  sin x
1  sin x
(1  sin x )  (1  sin x )
 2 sin x
1
sin x
 2 *
*
R
=
=
2
2
2
2
cos x cos x
cos x
cos x
cos x
cos x
6. Verify the identity:
sin(4x) = 4sinxcosx – 8sin3xcosx
L = sin(4x) = 2sin2xcos2x = 2(2sinxcosx)(cos2x - sin2x) = 4sinxcosx(cos2x - sin2x)
= 4sinxcos3x - 4sin3xcosx
= 4sinx*cosx*cos2x - 4sin3xcosx
= 4sinx*cosx*(1 - sin2x) - 4sin3xcosx
= 4sinxcosx(1 - sin2x) - 4sin3xcosx
= 4sinxcosx -4 sin3xcosx - 4sin3xcosx = R
7. Verify the identity: sin2x(1 + cotx) + cos2x(1 - tanx) + cot2x = csc2x
L = sin2x + sin2x cotx + cos2x - cos2x tanx + cot2x
= sin2x + sin2x(
= sin2x + sin2x(
cos x
sin x
) + cos2x - cos2x
+ cot2x
sin x
cos x
cos x
sin x
) + cos2x - cos2x
+ cot2x
sin x
cos x
= sin2x + sinxcosx + cos2x - cosxsinx + cot2x
= 1 + cot2x = csc2x = R
8. Show that the following equation is not an identity: sinx - tanx = 0
Try x = 30⁰: sin30⁰ - tan30⁰ =
9. Verify the identity:
1
3
which is not 0.
2
3
cot 2 x  1
= 1 - 2sin2x
2
1  cot x
cos 2 x
cos 2 x

1
1
2
sin x sin 2 x
cot x  1
cos 2 x  sin 2 x cos 2 x  sin 2 x
sin 2 x
L=
=
=
=
=
*

1
cos 2 x
sin 2 x
cos 2 x
1  cot 2 x
sin 2 x  cos 2 x
1
1
sin 2 x
sin 2 x
2
= cos2x - sin2x = (1 - sin2x) - sin2x = 1 - 2sin2x = R
10. Verify the identity:
L=
sin( x  y )
cos x cos y
 tan x  tan y
sin( x  y ) sin x cos y  sin y cos x
sin x cos y sin y cos x
=
=

 tan x  tan y  R
cos x cos y
cos x cos y
cos x cos y cos x cos y
11. Verify the identity:
2 cos 2 x
= cotx - tanx
sin 2 x
2 cos 2 x
=
sin 2 x
2(cos 2 x  sin 2 x) cos 2 x  sin 2 x
cos 2 x
sin 2 x
cos x sin x





 cot x  tan x
2 sin x cos x
sin x cos x
sin x cos x sin x cos x sin x cos x
L=
12. Verify the identity: tanx + cotx = 2 csc2x
R = 2 csc2x =2*
1
1
2
1


= 2*
sin 2 x
2 sin x cos x 2 sin x cos x sin x cos x
sin x
cos x
1
sin 2 x
cos 2 x
sin 2 x  cos 2 x
+
=
+
=
=
cos x
sin x
sin x cos x
sin x cos x
sin x cos x
sin x cos x
Since both sides simplify to the same quantity this verifies that the original equation is
an identity.
L = tanx + cotx =