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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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)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

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