5 - Finding Reference Angles
Transcript of 5 - Finding Reference Angles
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Finding Reference Angles
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It is necessary to be able to make larger angles smaller.
We do this by finding reference angles:
Step:
1. Start by drawing the given angle
2. Now, we determine how many
degrees it is ntil we get to the
!"a!is #hori$on%
&. 'his is yor reference angle(. ) reference angle is always
positive
Determine the reference anglefor 140 degree angle.
40 degrees is the reference angle
for 140 degrees
) reference angle mst be an
)*+' angle--
°140
°40
40140180 =−
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Determine the reference angle
for a 240 degree angle.
Determine the reference angle
for a 323 degree angle.
In the first adrant, the
angles are acte, so no need
to find a reference angle.
°240
°60
60180240 =−
°323
37323360 =−
°37
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Determine the reference angle
for a 470 degree angle.
Determine the reference angle
for a -125 degree angle.
110360470 =−
70110180 =−
°470
°70
°−125
55125180 =−
°55
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Homewor
!age " # $
%1&'2"
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)pplications of reference angles:
!press as a fnction of
a positive acte angle:
Step:
1. Start by drawing the given angle2. /ind the reference angle
&. 0ewrite the fnction sing reference
angle
(. etermine SIN of fnction in
adrant where it was drawn
(hat is tan in )*adrant ++,
A
/
°100tan
80100180 =−
°100
°80
°80tan
°− 80tan
°−=° 80tan100tan
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!press as a fnction of
a positive acte angle:!press as a fnction of
a positive acte angle:
°241cos
°241
61180241 =−
°61
°61cos
°− 61cos
°492sin
°492
132360492 =−
48132180 =−
°48
°48sin
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)pplications of reference angles:
/ind the e!act vale of
each e!pression:
Step:
1. Start by drawing the given angle
2. /ind the reference angle
&. 0ewrite the fnction sing reference
angle
(. etermine SIN of fnction in
adrant where it was drawn
3. Now find e!act vale sing e!act
vale chart
Reference
angle
*os in ad II is negative
Since cos is negative inadrant II, then the e!act
vale is negative as well
0
2
1
2
2
2
31
12
3
2
22
10
03
31 3 undefined
°120cos
°120
60120180 =−
°60
°60cos
°− 60cos2
1−=
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/ind the e!act vale of each e!pression:
°315tan
°315
45315360 =−
°45
°45tan
°− 45tan
02
1
2
2
2
31
12
3
2
2
2
10
03
31 3 undefined
1−=
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/ind the e!act vale of each e!pression:
°120sin
°120
°60
60120180 =−
°60sin2
3=
°210cos
°210
30180210 =−
°30
°30cos
°− 30cos2
3−=
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/ind the e!act vale of each e!pression:
°600sin
°600
240360600 =−
60180240 =−
°60
°60sin
°− 60sin2
3−=
( )°−30tan
°−30
°30
°30tan
°− 30tan3
3−=
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!act 4ales of adrantal angles:
(e are going to grah fo*r oints and *se
o*r nowledge of the *nit circle to hel *s
*st *se these oints to find the sin or cos
of an angle falling on that line.
y"vale is sin
θ cos
θ sin
( )θ θ sin,cos P
( )0,1
( )1,0
( )0,1−
( )1,0 −=90sin 1
=180cos 1−
=360sin 0
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!act 4ales of adrantal angles:
How do we find tan of these angles,Rememer
*ndefined
*ndefined
( )0,1
( )1,0
( )0,1−
( )1,0 −
θ
θ θ
cos
sintan =
=90tan =90cos
90sin=
0
1
=180tan =180cos
180sin=
−1
00
=270tan
=360tan 0