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Dr. Neal, WKU MATH 117 Arithmetic Formulas Given angles A and B , there several important arithmetic formulas for the sine and cosine of the angles A ± B , as well as angles such as ! A , 90º !B , 2A , and B / 2 . In each case, we can obtain an exact formula for the result in terms of the sine and cosine of the original angles A and B without knowing the actual values of the angles A and B .
Negative Angles
A
! A
(x , y)
(x ,! y)
Because angles A and ! A create the same x -coordinates on the unit circle, but negative y -coordinates, we have
cos A = cos(!A)
and
sin(!A) = ! sin A
Complimentary Angles
A
(x , y)
(y, x )
A
90º ! A
Because the x and y values are switched for complimentary angles such as 30º and 60º, we have
sin(90 º! A) = cos(A)
and
cos(90º ! A) = sin(A)
Supplementary Angles
AA
180º ! A(x , y)(!x, y)
Because supplementary angles, such as 30º and 150º, have the same y -coordinates but negative x -coordinates, we have
sin(A) = sin(180º !A)
and
cos(180º ! A) = ! cos(A)
Dr. Neal, WKU
Addition/Subtraction Formulas
sin(A + B) = sin A cosB + cosA sinB cos(A + B) = cosA cosB ! sinA sinB
sin(A ! B) = sin A cosB ! cosA sinB cos(A ! B) = cosA cosB + sinA sinB
Double Angle Formulas
sin(2A) = 2 sinA cosA cos(2A) = cos2 A ! sin2 A
Half-Angle Formulas
sinB
2
!
" #
$
% & = +
1' cosB
2 cos B
2
!
" #
$
% & = ±
1+ cosB
2
Example. Assume tan A = 27
, with angle A in Quadrant III, and csc B = ! 54
, with angle B in Quadrant IV. (a) Use right-triangle trig to determine the sines and cosines of angles A and B . (b) Find sin(A + B) , cos(A + B) , sin(A ! B) , cos(A ! B) . (c) Determine what quadrants the angles A + B and A ! B lie in. (d) Find sin(2A) and cos(2A) . Determine what quadrant the angle 2A lies in. (e) Find sin(B / 2) and cos(B / 2) . Determine what quadrant the angle B / 2 lies in. Solution. With A in III and tan A = 2
7, both x and y are negative. So y = !2 , x = !7 ,
and z = 22+ 7
2= 53 . For B in IV, y is negative but x is positive. Also
sin B =1
csc B= !
4
5.
–7
–253
Angle A in Quadrant III
!45
52! 4
2= 3
Angle B in Quadrant IV
sin A = !2
53 and cos A = !
7
53 sin B = !
4
5 and cos B = 3
5
Dr. Neal, WKU
sin(A + B) = sin A cosB + cos A sinB
=!2
53"3
5+
!7
53"!4
5
=!6 + 28
5 53=
22
5 53
cos(A + B) = cos A cosB ! sin A sinB
=!7
53"3
5!!2
53"!4
5
=!21 ! 8
5 53=
!29
5 53
sin(A ! B) = sin A cosB ! cos A sinB
=!2
53"3
5!
!7
53"!4
5
=!6 ! 28
5 53=
!34
5 53
cos(A ! B) = cosA cosB + sin A sinB
=!7
53"3
5+!2
53"!4
5
=!21 + 8
15 53=
!13
5 53
A + B must be in Quadrant II (cos is neg, sin is pos.) A ! B must be in Quadrant III (sin, cos are both neg.)
sin(2A) = 2 sinA cosA
= 2!2
53
"
# $
%
& '
!7
53
"
# $
%
& '
=28
53
cos(2A) = cos2 A ! sin2 A
=!7
53
"
# $
%
& ' 2
!!2
53
"
# $
%
& ' 2
=49
53!4
53=45
53
2A must be in the 1st Quadrant, because both sin and cos are positive.
For angle B / 2 , first note that angle B is in Quadrant IV. Thus, 270º! B ! 360º . Dividing by 2, we obtain: 135º ! B
2!180º . Thus, B / 2 is in Quadrant II. So the sine will
be positive, but the cosine will be negative.
sinB
2
! " #
$ % & = +
1 ' cosB
2=
1 '3
5
2
=2
5(1
2=
1
5=
5
5
cos
B
2
! " #
$ % & = '
1 + cos B
2= '
1 +3
5
2
= '8
5(1
2= '
4
5= '
2
5= '
2 5
5
Dr. Neal, WKU
15º and 75º Angles
The 15º and 75º reference angles througout the quadrants create many examples for using the Sum and Difference Angle Formulas. Each angle can be written as a sum or difference involving the 30º, 45º, and 60º reference angles for which we know the sines and cosines. For example, we can use 15º = 60º – 45º or 15º = 45º – 30º.
15º
75º105º
165º
195º 345º
255º 285º Example. Compute the exact value of sin(15º ) and cos(15º ) . Then compute sin(75º ) and cos(75º ) . Solution. Using 15º = 45º – 30º, we have sin(15º) = sin(45º!30º )
= sin 45º cos30º !cos45º sin30º
=2
2"
3
2!
2
2"1
2
=6 ! 2
4
and
cos(15º) = cos(45º!30º )
= cos 45º cos30º+ sin 45º sin30º
=2
2"
3
2+
2
2"1
2
=6 + 2
4
Because 75º is the complement of 15º, we should have that sin(75º) = cos(15º) and cos(75º) = sin(15º) . But using 75º = 45º + 30º, we obtain sin(75º ) = sin(45º +30º )
= sin 45º cos 30º +cos 45º sin 30º
=2
2!
3
2+
2
2!1
2
=6 + 2
4= cos(15º )
and
cos(75º ) = cos(45º +30º )
= cos 45º cos 30º ! sin 45º sin 30º
=2
2"
3
2!
2
2"1
2
=6 ! 2
4= sin(15º)
Dr. Neal, WKU
Exercises
1. Suppose sec(A) = ! 54
with A in Quad II, and cot(B) = 710
with B in Quad III. Draw the information on right triangles. Then find the exact values of
(a) sin(A + B) and sin(A ! B) (b) cos(A + B) and cos(A ! B)
(c) sin(2B) and cos(2B) (d) sin A
2
!
" #
$
% & and cos A
2
!
" #
$
% &
(e) Explain what quadrants the angles A + B , A ! B , 2B , and A / 2 are in.
2. Use the sum/addition formulas to find the exact values of (a) sin 105º( ) and cos 105º( ) (b) sin 165º( ) and cos 165º( ) (c) sin 255º( ) and cos 255º( ) (d) sin 345º( ) and cos 345º( )
Dr. Neal, WKU
Answer
5
3
-4
Angle A in II
–7
–10
149
Angle B in III
sin(A + B) = sin A cosB + cos A sinB
=3
5!
"7
149+"4
5!"10
149
="21 + 40
5 149=
19
5 149
cos(A + B) = cos A cosB ! sin A sinB
=!4
5"
!7
149!3
5"
!10
149
=28 + 30
5 149=
58
5 149
sin(A ! B) = sin A cosB ! cos A sinB
=3
5"
!7
149!!4
5"
!10
149
=!21 ! 40
5 149=
!61
5 149
cos(A ! B) = cosA cosB + sin A sinB
=!4
5"
!7
149+3
5"
!10
149
=28 ! 30
5 149=
!2
5 149
A + B must be in Quadrant I (cos is pos, sin is pos.) A ! B must be in Quadrant III (sin, cos are both neg.)
sin(2B) = 2sinB cosB
= 2!10
149
"
# $
%
& '
!7
149
"
# $
%
& '
=140
149
cos(2B) = cos2 B ! sin2 B
=!7
149
"
# $
%
& ' 2
!!10
149
"
# $
%
& ' 2
=49
149!100
149= !
51
149
2B must be in Quadrant I!, because cos is neg and sin is pos.
For angle A / 2 , first note that angle A is in Quadrant II. Thus, 90º! A ! 180º . Dividing by 2, we obtain: 45º ! A
2! 90º . Thus, A / 2 is in Quadrant I. So the sine will be
positive and the cosine will be positive.
sinA
2
!
" #
$
% & = +
1 ' cosA
2=
1''4
5
2
=9
5(1
2=
9
10=
3
10
cos
A
2
!
" #
$
% & = '
1 + cosA
2=
1+'4
5
2
=1
5(1
2=
1
10=1
10
