Asked by Tiffany Enlow
4. Find the exact value for sin(x+y) if sinx=-4/5 and cos y = 15/17. Angles x and y are in the fourth quadrant.
5. Find the exact value for cos 165degrees using the half-angle identity.
1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0 less than or equal to x <2pi.
2. Solve: 2 sinx - 1 = 0 for 0degrees less than or equal to x <360degrees.
3. Solve: sin^2x = cos^2x for 0degrees is less than or equal to x < 360degrees.
4. Solve sinx - 2sinx cosx = 0 for 0 is less than or equal to x < 2pi.
5. Find the exact value for cos 165degrees using the half-angle identity.
1. Solve: 2 cos^2x - 3 cosx + 1 = 0 for 0 less than or equal to x <2pi.
2. Solve: 2 sinx - 1 = 0 for 0degrees less than or equal to x <360degrees.
3. Solve: sin^2x = cos^2x for 0degrees is less than or equal to x < 360degrees.
4. Solve sinx - 2sinx cosx = 0 for 0 is less than or equal to x < 2pi.
Answers
Answered by
Reiny
#4
sin(x+y) = sinx cosy + cosx siny
so we need the cosx and the siny
from sinx = -4/5 we know we are dealing with the 3,4,5 right-angled triangle, so in the fourth quadrant cosx = 3/5
from cosy = 15/17 we know we are dealing with the 8,15,17 right-angled triangle
so in the fourth quadrant, siny = -8/17
then
sin(x+y) = sinx cosy + cosx siny
= (-4/5)(15/17) + (3/5)(-8/17)
= -84/85
sin(x+y) = sinx cosy + cosx siny
so we need the cosx and the siny
from sinx = -4/5 we know we are dealing with the 3,4,5 right-angled triangle, so in the fourth quadrant cosx = 3/5
from cosy = 15/17 we know we are dealing with the 8,15,17 right-angled triangle
so in the fourth quadrant, siny = -8/17
then
sin(x+y) = sinx cosy + cosx siny
= (-4/5)(15/17) + (3/5)(-8/17)
= -84/85
Answered by
Reiny
cos 2A = 2cos^2 A - 1
let A = 165, then 2A = 330
so let's find cos 330
cos(330)
= cos(360-30)
= cos360 cos30 + sin360 sin30
= (1)(√3/2) + (0)(1/2) = √3/2
then √3/2 = 2cos^2 165 - 1
(√3 + 2)/4 = cos^2 165
cos 165 = -√(√3 + 2))/2
(algebraically, our answer would have been ± , but I picked the negative answer since 165 is in the second quadrant, and the cosine is negative in the second quadrant)
let A = 165, then 2A = 330
so let's find cos 330
cos(330)
= cos(360-30)
= cos360 cos30 + sin360 sin30
= (1)(√3/2) + (0)(1/2) = √3/2
then √3/2 = 2cos^2 165 - 1
(√3 + 2)/4 = cos^2 165
cos 165 = -√(√3 + 2))/2
(algebraically, our answer would have been ± , but I picked the negative answer since 165 is in the second quadrant, and the cosine is negative in the second quadrant)
Answered by
Reiny
I will give you hints for the rest,
you do them, and let me know what you get.
#1 factor it as
(2cosx - 1)(cosx -1) = 0
so cosx - 1/2 or cosx = 1
take it from there, you should get 3 answers.
#2, the easiest one
take the 1 to the other side, then divide by 2,
#3, take √ of both sides to get
sin 2x = ± cos 2x
sin2x/cos2x = ± 1
tan 2x = ± 1 etc
#4 factor out a sinx etc
you do them, and let me know what you get.
#1 factor it as
(2cosx - 1)(cosx -1) = 0
so cosx - 1/2 or cosx = 1
take it from there, you should get 3 answers.
#2, the easiest one
take the 1 to the other side, then divide by 2,
#3, take √ of both sides to get
sin 2x = ± cos 2x
sin2x/cos2x = ± 1
tan 2x = ± 1 etc
#4 factor out a sinx etc
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