Value of this trigonometric expression

[songoku]

Homework Statement:

Given that sin A + sin B = 1 and cos A + cos B = 0, find the value of 12 cos 2A + 4 cos 2B

Relevant Equations:

Trigonometry

Let:
equation 1 : sin A + sin B = 1
equation 2 : cos A + cos B = 0

Squaring both sides of equation 1 and 2 then add the result gives me: cos (A - B) = -1/2

Then how to proceed? Thanks

 

[anuttarasammyak]

We know from equation 1
$$0<A,B<\pi$$
so equation 2 says
$$A,B=\frac{\pi}{2}-\theta, \frac{\pi}{2}+\theta$$ where
$$0<\theta<\frac{\pi}{2}$$

 

[songoku]

We know from equation 1
$$0<A,B<\pi$$
so equation 2 says
$$A,B=\frac{\pi}{2}-\theta, \frac{\pi}{2}+\theta$$ where
$$0<\theta<\frac{\pi}{2}$$

I get it. From there, we can get the value of A and B.

Thank you very much anuttarasammyak

 

[Mark44]

We know from equation 1
$$0<A,B<\pi$$

Edited...
No, this isn't true. Consider $A = \frac \pi 6$ and $B = \frac{13\pi} 6$. Then $\sin A + \sin B = 1$. This is just one of an infinite number of counterexamples.

so equation 2 says
$$A,B=\frac{\pi}{2}-\theta, \frac{\pi}{2}+\theta$$ where
$$0<\theta<\frac{\pi}{2}$$

I've tried several different approaches, but none has been successful so far.

 

[songoku]

Edit: B should read $\frac{13\pi} 6$.
No, this isn't true. Consider $A = \frac \pi 6$ and $B = \frac{7\pi} 6$. Then $\sin A + \sin B = 1$.

I am not sure I understand your hint, but:
sin (π/6) + sin (7π/6) = sin (π/6) + sin (π + π/6) = sin (π/6) - sin (π/6) = 0 ?

Thanks

 

[Delta2]

Hint: First focus to prove that $\cos 2A=\cos 2B$. To prove this you can proceed relatively easily from $\cos A=-\cos B$ by squaring both sides.

Once you prove that $\cos 2A=\cos 2B$ then you can solve this relatively simple trigonometric equation and relate directly the angles A,B. Then use equation 1.

 

[Mark44]

I am not sure I understand your hint, but:
sin (π/6) + sin (7π/6) = sin (π/6) + sin (π + π/6) = sin (π/6) - sin (π/6) = 0 ?

Thanks

I have edited my earlier post.
What I meant was $A = \frac \pi 6$ and $B = \frac {13\pi} 6$. Then $\sin A + \sin B = 1$, but it's not true that both A and B are in the interval $[0, \frac \pi 2]$ as was earlier claimed.

 

[ehild]

Homework Statement:: Given that sin A + sin B = 1 and cos A + cos B = 0, find the value of 12 cos 2A + 4 cos 2B
Relevant Equations:: Trigonometry

Let:
equation 1 : sin A + sin B = 1
equation 2 : cos A + cos B = 0

eq. 2 says that cosB=-cosA, From eq.1, it follows that both angles are in the interval (0,pi). Therefore B=pi-A. Substitute for B into eq. 1
sinA+sin(pi-A)=2sinA=1--->sinA=0.5 , A=30°, B=°150°.

 

[Delta2]

eq. 2 says that cosB=-cosA, From eq.1, it follows that both angles are in the interval (0,pi). Therefore B=pi-A. Substitute for B into eq. 1
sinA+sin(pi-A)=2sinA=1--->sinA=0.5 , A=30°, B=°150°.

I don't understand how you conclude that the angles are in the interval $(0,\pi)$. My approach at the very end of day concludes the same result, that is $\sin A=\sin B=0.5$ but without making any conclusions on what are the exact values of the angles.

 

[songoku]

I have edited my earlier post.
What I meant was $A = \frac \pi 6$ and $B = \frac {13\pi} 6$. Then $\sin A + \sin B = 1$, but it's not true that both A and B are in the interval $[0, \frac \pi 2]$ as was earlier claimed.

Sorry I still not fully understand. Isn't 13π/6 located in first quadrant? And I think post #2 claimed that A and B are in interval (0, π), not [0, π/2]

I don't understand how you conclude that the angles are in the interval $(0,\pi)$. My approach at the very end of day concludes the same result, that is $\sin A=\sin B=0.5$ but without making any conclusions on what are the exact values of the angles.

This is what I concluded from hint in post#2:

From sin A + sin B = 1, both sin A and sin B must be positive because if either one is negative, the value of the other trigonometry function will more than 1 (which is impossible) so this means 0 < A , B < π

 

[ehild]

My approach at the very end of day concludes the same result, that is $\sin A=\sin B=0.5$ but without making any conclusions on what are the exact values of the angles.

Yes, you noted that cos(A) = - cos(B) which directly means B= pi-+A. And the sine of both angles must be positive, so B=pi-A.
You overcomplicated the solution a bit by squaring and working with double angles.

 

[haruspex]

A=30°, B=150°.

Or vice versa. But because we are only interested in cos(2A) and cos(2B), it makes no difference.
I have the feeling there is some quick way that avoids finding the angles, but I don't see it.

 

[Delta2]

Or vice versa. But because we are only interested in cos(2A) and cos(2B), it makes no difference.
I have the feeling there is some quick way that avoids finding the angles, but I don't see it.

The way i did it avoids finding the angles (i found only sines/cosines of angles/double angles) but its not quicker, its longer actually...

 

[Ibix]

I have the feeling there is some quick way that avoids finding the angles, but I don't see it.

I managed it by writing $\cos A=-\cos B$ and $\sin A=1-\sin B$ and squaring both sides of each. This gives you $\sin B$ in a couple of lines.

 

[kuruman]

Or vice versa. But because we are only interested in cos(2A) and cos(2B), it makes no difference.
I have the feeling there is some quick way that avoids finding the angles, but I don't see it.

Quick way: Note that $|\cos A|=| \cos B |$. Therefore $|\sin A|=\sqrt{1-\cos^2 A}=\sqrt{1-\cos^2B}=|\sin B|$. Since the sum of the sines is equal to +1, both of them must be positive and $\sin A=\sin B=\frac{1}{2}$. Now $$\cos(2A)=1-2\sin^2A=1-2\left(\frac{1}{2}\right)^2=\frac{1}{2}=1-2\sin^2B=\cos(2B).$$