Sum and Difference Problem

[Jason76]

EDITED

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

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

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

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

Find angle A and B:

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

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

Now find rest:

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

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

Now back to the problem:

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

\(\displaystyle \dfrac{1 + 4\sqrt{3}}{6}\) Final answer

 

[DrPhil]

Given... \(\displaystyle \sin(A) = -\dfrac{1}{2}\)
and ...\(\displaystyle \pi \lt A \lt \dfrac{3\pi}{2}\)
and ...\(\displaystyle -\cos(B) = \dfrac{1}{4}\) in quadrant IV

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

ok - I have gussied up the LaTeX a little, but Where Is Your Work?
You can find \(\displaystyle \cos A\) from the sine, and \(\displaystyle \sin B\) 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 \(\displaystyle \sin^2 A + \cos^2 A = 1\) to sind the two missing functions. Rather than trying to find arccos(1/4), much better to use the identity

\(\displaystyle \sin(A - B) = \sin A\ \cos B - \cos A\ \sin B\)

 

[Jason76]

ok - I have gussied up the LaTeX a little, but Where Is Your Work?
You can find \(\displaystyle \cos A\) from the sine, and \(\displaystyle \sin B\) 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 \(\displaystyle \sin^2 A + \cos^2 A = 1\) to sind the two missing functions. Rather than trying to find arccos(1/4), much better to use the identity

\(\displaystyle \sin(A - B) = \sin A\ \cos B - \cos A\ \sin B\)

All done and edited, but might not be correct.

 

[soroban]

Hello, Jason76!

There is a serious error in the problem.

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

. . . . . . . \(\displaystyle \cos B = \text{-}\tfrac{1}{4},\:B \in \text{Quad IV}\) . ??

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

cosine is positive in Quadrant IV.

 

[Jason76]

Double Angle Problem

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

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

with formula:

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

Find angle B

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

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

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

 

[Jason76]

Hello, Jason76!

There is a serious error in the problem.


cosine is positive in Quadrant IV.

So do we find a reference angle and work with that, something like that?

 

[DrPhil]

EDITED

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

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

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

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

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

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

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

Now find rest:

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

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

Now back to the problem:

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

\(\displaystyle \dfrac{1 + 4\sqrt{3}}{6}\) Final answer

You DO have to check the statement of the problem, because cosB is positive in quadrant IV.

 

[DrPhil]

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

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

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

with formula:

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

Find angle B NOT RELEVANT!!

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

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

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

.

 

[srmichael]

EDITED

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

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

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

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

Find angle A and B:

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

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

Now find rest:

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

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

Now back to the problem:

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

\(\displaystyle \dfrac{1 + 4\sqrt{3}}{6}\) Final answer

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

 

[Deleted member 4993]

EDITED

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

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

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

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.