Value of this trigonometric expression

• songoku
In summary, sin A + sin B = 1 and cos A + cos B = 0. When squared, cos(A-B) = -1/2. To solve for A and B, use equations 1 and 2: A=30°, B=150°.
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

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
anuttarasammyak said:
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

anuttarasammyak said:
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.
anuttarasammyak said:
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.

Last edited:
Mark44 said:
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

Last edited by a moderator:
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.

Last edited:
songoku
songoku said:
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.

songoku said:
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°.

songoku and Delta2
ehild said:
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
Mark44 said:
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]

Delta2 said:
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

Delta2
Delta2 said:
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.

songoku and Delta2
ehild said:
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.

songoku
haruspex said:
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...

songoku and Ibix
haruspex said:
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.

songoku, etotheipi and Delta2
haruspex said:
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 $$\cos(2A)=1-2\sin^2A=1-2\left(\frac{1}{2}\right)^2=\frac{1}{2}=1-2\sin^2B=\cos(2B).$$

songoku and Delta2

1. What is the purpose of a trigonometric expression?

A trigonometric expression is used to represent relationships between angles and sides in a right triangle. It is often used in mathematics and science to solve problems involving angles and distances.

2. How do you find the value of a trigonometric expression?

The value of a trigonometric expression can be found by using a calculator or by using trigonometric identities and formulas to simplify the expression. It is important to remember to use the correct units (degrees or radians) when finding the value.

3. Can a trigonometric expression have multiple values?

Yes, a trigonometric expression can have multiple values depending on the given angle and the type of trigonometric function used. For example, the sine function can have two values for a given angle, while the tangent function can have infinitely many values.

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The most common types of trigonometric expressions are sine, cosine, tangent, cotangent, secant, and cosecant. These functions are used to represent the ratios of sides in a right triangle and can be used to solve various types of problems involving angles and distances.

5. How are trigonometric expressions used in real life?

Trigonometric expressions are used in real life in fields such as engineering, physics, and astronomy. They are used to calculate distances, heights, and angles in various structures and objects. They are also used in navigation and mapping to determine locations and distances between points.

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