Sum and Difference Problem

Jason76

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EDITED

Given sin(A)=12\displaystyle \sin(A) = -\dfrac{1}{2} and π\displaystyle \pi is less than a\displaystyle a is less than 3π2\displaystyle \dfrac{3\pi}{2} and cos(B)=14\displaystyle \cos(B) = -\dfrac{1}{4} in quadrant IV\displaystyle IV

Find: sin(AB)\displaystyle \sin(A - B)

sin(AB)=[sin(A)][cos(B)]+[cos(A)][sin(B)]\displaystyle \sin(A - B) = [\sin(A)][\cos(B)] + [\cos(A)][\sin(B)]

12(14)+[cos(A)][(sin(B)]\displaystyle -\dfrac{1}{2}(-\dfrac{1}{4}) + [\cos(A)][(\sin(B)]

Find angle A and B:

arcsin(12)=30=A\displaystyle \arcsin (-\dfrac{1}{2}) = -30^{\circ} = A

arccos(14)=104.5=B\displaystyle \arccos( -\dfrac{1}{4}) = 104.5^{\circ} = B

Now find rest:

cos(30)=32\displaystyle \cos(-30^{\circ} ) = \dfrac{\sqrt{3}}{2}

sin(104.5)=1\displaystyle \sin(104.5^{\circ} ) = 1

Now back to the problem:

12(14)+(32)(1)\displaystyle -\dfrac{1}{2}(-\dfrac{1}{4}) + (\dfrac{\sqrt{3}}{2})(1)

1+436\displaystyle \dfrac{1 + 4\sqrt{3}}{6} Final answer :confused:
 
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Given... sin(A)=12\displaystyle \sin(A) = -\dfrac{1}{2}
and ...π<A<3π2\displaystyle \pi \lt A \lt \dfrac{3\pi}{2}
and ...cos(B)=14\displaystyle -\cos(B) = \dfrac{1}{4} in quadrant IV

Find: sin(AB)\displaystyle \sin(A - B)
ok - I have gussied up the LaTeX a little, but Where Is Your Work?
You can find cosA\displaystyle \cos A from the sine, and sinB\displaystyle \sin B from the cosine. and then use the formula for sine of the difference. Where are you getting stuck?

EDIT - ok, there is some work shown in YOUR edit!

Use the relation sin2A+cos2A=1\displaystyle \sin^2 A + \cos^2 A = 1 to sind the two missing functions. Rather than trying to find arccos(1/4), much better to use the identity

sin(AB)=sinA cosBcosA sinB\displaystyle \sin(A - B) = \sin A\ \cos B - \cos A\ \sin B
 
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ok - I have gussied up the LaTeX a little, but Where Is Your Work?
You can find cosA\displaystyle \cos A from the sine, and sinB\displaystyle \sin B from the cosine. and then use the formula for sine of the difference. Where are you getting stuck?

EDIT - ok, there is some work shown in YOUR edit!

Use the relation sin2A+cos2A=1\displaystyle \sin^2 A + \cos^2 A = 1 to sind the two missing functions. Rather than trying to find arccos(1/4), much better to use the identity

sin(AB)=sinA cosBcosA sinB\displaystyle \sin(A - B) = \sin A\ \cos B - \cos A\ \sin B

All done and edited, but might not be correct.
 
Hello, Jason76!

There is a serious error in the problem.


Given: sinA=-12,A Quad III\displaystyle \text{Given: }\:\sin A = \text{-}\tfrac{1}{2},\:A \in \text{ Quad III}

. . . . . . . cosB=-14,BQuad IV\displaystyle \cos B = \text{-}\tfrac{1}{4},\:B \in \text{Quad IV} . ??

Find: sin(AB)\displaystyle \text{Find: }\,\sin(A - B)

cosine is positive in Quadrant IV.
 
Double Angle Problem

Given sin(A)=12\displaystyle \sin(A) = -\dfrac{1}{2} and π\displaystyle \pi is less than a\displaystyle a is less than 3π2\displaystyle \dfrac{3\pi}{2} and cos(B)=14\displaystyle \cos(B) = -\dfrac{1}{4} in quadrant IV\displaystyle IV

cos(2B)\displaystyle \cos(2B)

with formula:

cos(2θ)=2cos2θ1\displaystyle \cos(2\theta) = 2\cos^{2}\theta - 1

Find angle B

arccos(14)=104.5\displaystyle \arccos(-\dfrac{1}{4}) = 104.5^{\circ}

cos(2(104.5))=2cos2(104.5)1\displaystyle \cos(2(104.5^{\circ})) = 2\cos^{2}(104.5^{\circ}) - 1

2cos2(104.5)1=.875\displaystyle 2\cos^{2}(104.5^{\circ}) - 1 = -.875
 
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EDITED

Given sin(A)=12\displaystyle \sin(A) = -\dfrac{1}{2} and π\displaystyle \pi is less than a\displaystyle a is less than 3π2\displaystyle \dfrac{3\pi}{2} and cos(B)=14\displaystyle \cos(B) = -\dfrac{1}{4} in quadrant IV\displaystyle IV

Find: sin(AB)\displaystyle \sin(A - B)

sin(AB)=[sin(A)][cos(B)]+[cos(A)][sin(B)]\displaystyle \sin(A - B) = [\sin(A)][\cos(B)] + [\cos(A)][\sin(B)]

12(14)+[cos(A)][(sin(B)]\displaystyle -\dfrac{1}{2}(-\dfrac{1}{4}) + [\cos(A)][(\sin(B)]

Find angle A and B: NO, don't find A and B, rather find cosA and sinB

arcsin(12)=30=A\displaystyle \arcsin (-\dfrac{1}{2}) = -30^{\circ} = A NO, A is in quadrant III...

arccos(14)=104.5=B\displaystyle \arccos( -\dfrac{1}{4}) = 104.5^{\circ} = B NOT clear what quadrant B is in

Now find rest:

cos(30)=32\displaystyle \cos(-30^{\circ} ) = \dfrac{\sqrt{3}}{2} NO, cosA = -sqrt[1 - (1/2)^2] is negative

sin(104.5)=1\displaystyle \sin(104.5^{\circ} ) = 1NO, CAN'T be 1!! sinB = ±sqrt[1 - (1/4)^2], sign depends on quadrant

Now back to the problem:

12(14)+(32)(1)\displaystyle -\dfrac{1}{2}(-\dfrac{1}{4}) + (\dfrac{\sqrt{3}}{2})(1)

1+436\displaystyle \dfrac{1 + 4\sqrt{3}}{6} Final answer :confused:
You DO have to check the statement of the problem, because cosB is positive in quadrant IV.
 
Given sin(A)=12\displaystyle \sin(A) = -\dfrac{1}{2} and π\displaystyle \pi is less than a\displaystyle a is less than 3π2\displaystyle \dfrac{3\pi}{2} and cos(B)=14\displaystyle \cos(B) = -\dfrac{1}{4} in quadrant IV\displaystyle IV

B can't be in quadrant 4 if cosB is negative

cos(2B)\displaystyle \cos(2B)

with formula:

cos(2θ)=2cos2θ1\displaystyle \cos(2\theta) = 2\cos^{2}\theta - 1

Find angle B NOT RELEVANT!!

arccos(14)=104.5\displaystyle \arccos(-\dfrac{1}{4}) = 104.5^{\circ}

cos(2(104.5))=2cos2(104.5)1\displaystyle \cos(2(104.5^{\circ})) = 2\cos^{2}(104.5^{\circ}) - 1

2cos2(104.5)1=.875\displaystyle 2\cos^{2}(104.5^{\circ}) - 1 = -.875 2 cos^2(B) - 1 = 2(1/16) - 1 = -7/8
.
 
EDITED

Given sin(A)=12\displaystyle \sin(A) = -\dfrac{1}{2} and π\displaystyle \pi is less than a\displaystyle a is less than 3π2\displaystyle \dfrac{3\pi}{2} and cos(B)=14\displaystyle \cos(B) = -\dfrac{1}{4} in quadrant IV\displaystyle IV

Find: sin(AB)\displaystyle \sin(A - B)

\(\displaystyle \sin(A - B) = [\sin(A)][\cos(B)] + [\cos(A)][\sin(B)]\) ===> INCORRECT. Sin(A - B) = SinACosB - CosASinB

12(14)+[cos(A)][(sin(B)]\displaystyle -\dfrac{1}{2}(-\dfrac{1}{4}) + [\cos(A)][(\sin(B)]

Find angle A and B:

arcsin(12)=30=A\displaystyle \arcsin (-\dfrac{1}{2}) = -30^{\circ} = A

arccos(14)=104.5=B\displaystyle \arccos( -\dfrac{1}{4}) = 104.5^{\circ} = B

Now find rest:

cos(30)=32\displaystyle \cos(-30^{\circ} ) = \dfrac{\sqrt{3}}{2}

sin(104.5)=1\displaystyle \sin(104.5^{\circ} ) = 1

Now back to the problem:

12(14)+(32)(1)\displaystyle -\dfrac{1}{2}(-\dfrac{1}{4}) + (\dfrac{\sqrt{3}}{2})(1)

1+436\displaystyle \dfrac{1 + 4\sqrt{3}}{6} Final answer :confused:

Along with the other errors that were pointed out, add the one above too.
 
EDITED

Given sin(A)=12\displaystyle \sin(A) = -\dfrac{1}{2} and π\displaystyle \pi is less than a\displaystyle a is less than 3π2\displaystyle \dfrac{3\pi}{2} and

cos(B)=14\displaystyle \cos(B) = -\dfrac{1}{4} in quadrant IV\displaystyle IV

Find: sin(AB)\displaystyle \sin(A - B)

I assume - that means B is in IV quadrant.

cos(B) CANNOT be negative (<0) - when B is in 4th quadrant.

Fix your post - then we can help you get the correct answer.

If that is the "correct" statement of the problem, then,

the problem is describing an impossible situation - hence CANNOT be solved.
 
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