3 pi /2 < x < 2 pi
mean x lie in IV quadrant
tan ( x ) = + OR - sqrt [ 1 - cos ( x ) ^ 2 ] / cos ( x )
in this case:
tan ( x ) = + OR - sqrt ( 1 - 0.6 ^ 2 ) / 0.6
tan ( x ) = + OR - sqrt ( 1 - 0.36 ) / 0.6
tan ( x ) = + OR - sqrt ( 0.64 ) / 0.6
tan ( x ) = + OR - 0.8 / 0.6
tan ( x ) = + OR - 8 / 6
tan ( x ) = + OR - 2 * 4 / ( 2 * 3 )
tan ( x ) = + OR - 4 / 3
In IV quadrant tangent are negative so :
tan ( x ) = - 4 / 3
cos ( x ) = 0.6
sin ( x ) = + OR - sqrt [ 1 - cos ( x ) ^ 2 ]
sin ( x ) = + OR - sqrt ( 1 - 0.6 ^ 2 )
sin ( x ) = + OR - sqrt ( 1 - 0.36 )
sin ( x ) = + OR - sqrt ( 0.64 )
sin ( x ) = + OR - 0.8
In IV quadrant sin ( x ) are negative so :
sin ( x ) = - 0.8
sin ( x + pi ) = - sin ( x )
sin ( x + pi ) = - ( - 0.8 )
sin ( x + pi ) = 0.8
Find the tan(x), the sin (x + pi) if it is known that cos (x)=.6 and that (3pi)/2 less than x less than 2pi.
Find the sine (2x), the tan (2x), and the cos (x/2) if it is known that cos (x)=.6 and that (3pi)/2 less than x less than 2pi.
3 answers
sin ( x ) = + OR - sqrt [ 1 - cos ( x ) ^ 2 ]
sin ( x ) = + OR - sqrt ( 1 - 0.6 ^ 2 )
sin ( x ) = + OR - sqrt ( 1 - 0.36 )
sin ( x ) = + OR - sqrt ( 0.64 )
sin ( x ) = + OR - 0.8
In IV quadrant sin ( x ) are negative so :
sin ( x ) = - 0.8
sin ( 2 x ) = 2 * sin ( x ) * cos ( x )
sin ( 2 x ) = 2 * ( - 0.8 ) * ( 0.6 )
sin ( 2x ) = - 0.96
tan ( 2 x ) = [ 2 * sin ( x ) * cos ( x ) ] / [ cos ^ 2 ( x ) - sin ^ 2 ( x ) ]
[ 2 * sin ( x ) * cos ( x ) ] = - 0.96
cos ^ 2 ( x ) = 0.6 ^ 2 = 0.36
sin ^ 2 ( x ) = ( - 0.8 ) ^ 2 = 0.64
tan ( 2 x ) = - 0.96 / ( 0.36 - 0.64 )
tan ( 2 x ) = - 0.96 / - 0.28
tan ( 2 x ) = - 96 / 28
tan ( 2 x ) = 4 * 3 * 8 / ( 4 * 7 )
tan ( 2 x ) = 24 / 7
If 3 pi / 2 < x < 2 pi
then 3 pi / 4 < x / 2 < pi
mean x lie in III quadrant
[ cos ( x / 2 ) ] ^ 2 = [ cos( x ) + 1 ] / 2
cos ( x / 2 ) = + OR - sqrt [ [ cos ( x ) + 1 ] / 2 ]
cos ( x / 2 ) = + OR - sqrt [ ( 0.6 + 1 ) / 2 ]
cos ( x / 2 ) = + OR - sqrt ( 1.6 / 2 )
cos ( x / 2 ) ] = + OR - sqrt ( 0.8 )
cos ( x / 2 ) = + OR - sqrt ( 8 / 10 )
cos ( x / 2 ) = + OR - sqrt [ ( 2 * 4 / ( 2 * 5 ) ]
cos ( x / 2 ) = + OR - sqrt ( 4 / 5 )
cos ( x / 2 ) = + OR - 2 / sqrt ( 5 )
In III quadrant cosine are negative so :
cos ( x / 2 ) = - 2 / sqrt ( 5 )
sin ( x ) = + OR - sqrt ( 1 - 0.6 ^ 2 )
sin ( x ) = + OR - sqrt ( 1 - 0.36 )
sin ( x ) = + OR - sqrt ( 0.64 )
sin ( x ) = + OR - 0.8
In IV quadrant sin ( x ) are negative so :
sin ( x ) = - 0.8
sin ( 2 x ) = 2 * sin ( x ) * cos ( x )
sin ( 2 x ) = 2 * ( - 0.8 ) * ( 0.6 )
sin ( 2x ) = - 0.96
tan ( 2 x ) = [ 2 * sin ( x ) * cos ( x ) ] / [ cos ^ 2 ( x ) - sin ^ 2 ( x ) ]
[ 2 * sin ( x ) * cos ( x ) ] = - 0.96
cos ^ 2 ( x ) = 0.6 ^ 2 = 0.36
sin ^ 2 ( x ) = ( - 0.8 ) ^ 2 = 0.64
tan ( 2 x ) = - 0.96 / ( 0.36 - 0.64 )
tan ( 2 x ) = - 0.96 / - 0.28
tan ( 2 x ) = - 96 / 28
tan ( 2 x ) = 4 * 3 * 8 / ( 4 * 7 )
tan ( 2 x ) = 24 / 7
If 3 pi / 2 < x < 2 pi
then 3 pi / 4 < x / 2 < pi
mean x lie in III quadrant
[ cos ( x / 2 ) ] ^ 2 = [ cos( x ) + 1 ] / 2
cos ( x / 2 ) = + OR - sqrt [ [ cos ( x ) + 1 ] / 2 ]
cos ( x / 2 ) = + OR - sqrt [ ( 0.6 + 1 ) / 2 ]
cos ( x / 2 ) = + OR - sqrt ( 1.6 / 2 )
cos ( x / 2 ) ] = + OR - sqrt ( 0.8 )
cos ( x / 2 ) = + OR - sqrt ( 8 / 10 )
cos ( x / 2 ) = + OR - sqrt [ ( 2 * 4 / ( 2 * 5 ) ]
cos ( x / 2 ) = + OR - sqrt ( 4 / 5 )
cos ( x / 2 ) = + OR - 2 / sqrt ( 5 )
In III quadrant cosine are negative so :
cos ( x / 2 ) = - 2 / sqrt ( 5 )
first:
as Bosnian noted, x is in IV
given: cosx = .6 = 6/10 = 3/5
you should recognize the 3-4-5 right-angled triangle
so sinx = -4/5 ---> in IV the sine is negative
and tanx = -4/3 --->
tanx = sinx/cosx
= (-4/5)(3/5) = (-4/5)(5/3)
= -4/3
sin(x+π) = sinxcosπ + cosxsinπ
= (-4/5)(-1) + (.6)(0) = 4/5
2nd:
from above, x is the same angle, and
cosx = 3/5 and sinx = -4/5
sin(2x) = 2sinxcosx
= 2(-4/5)(3/5) = -24/25
tan(2x) = sin(2x)/cos(2x)
so we need cos(2x) = cos^2x - sin^2x
= 9/25 - 16/25 = -7/25
tan(2x) = (-24/25)/(-7/25) = 24/7
for cos(x/2) use cos 2A = 2cos^2 A - 1
cosx = 2cos^2 (x/2) - 1
3/5 = 2cos^2 (x/2) - 1
2cos^2 (x/2) = 8/5
cos^2 (x/2) = 8/10
cos (x/2) = ± √8/10)
but if x is in IV, then x/2 is in II, where the cosine is negative
so cos (x/2) = - √(8/10 )
= -√(4/5) = -2/√5
as Bosnian noted, x is in IV
given: cosx = .6 = 6/10 = 3/5
you should recognize the 3-4-5 right-angled triangle
so sinx = -4/5 ---> in IV the sine is negative
and tanx = -4/3 --->
tanx = sinx/cosx
= (-4/5)(3/5) = (-4/5)(5/3)
= -4/3
sin(x+π) = sinxcosπ + cosxsinπ
= (-4/5)(-1) + (.6)(0) = 4/5
2nd:
from above, x is the same angle, and
cosx = 3/5 and sinx = -4/5
sin(2x) = 2sinxcosx
= 2(-4/5)(3/5) = -24/25
tan(2x) = sin(2x)/cos(2x)
so we need cos(2x) = cos^2x - sin^2x
= 9/25 - 16/25 = -7/25
tan(2x) = (-24/25)/(-7/25) = 24/7
for cos(x/2) use cos 2A = 2cos^2 A - 1
cosx = 2cos^2 (x/2) - 1
3/5 = 2cos^2 (x/2) - 1
2cos^2 (x/2) = 8/5
cos^2 (x/2) = 8/10
cos (x/2) = ± √8/10)
but if x is in IV, then x/2 is in II, where the cosine is negative
so cos (x/2) = - √(8/10 )
= -√(4/5) = -2/√5
1 answer