Solve Trigonometric Ratios: 2ab/(b^2 - a^2)

In summary, the conversation discusses how to find the value of tan(x+y) given the equations sin x + sin y = a and cos x + cos y = b, and how to prove it to be equal to 2ab / (b^2-a^2). The solution involves using the sum and difference formulas for sines and cosines to rewrite the expressions and simplify them into the desired form. The expert summarizer provides additional guidance and corrections to help solve the problem.
  • #1
ritwik06
580
0

Homework Statement


Given:
sin x + sin y= a
cos x + cos y=b
find tan(x+y)
Prove it to be equal to"
[tex]\frac{2ab}{b^{2}-a^{2}}[/tex]


The Attempt at a Solution



I get stuck after this:
2ab= 2 sin (x+y)+sin 2x+sin 2y
[tex]b^{2}-a^{2}= cos 2x + cos 2y + 2 cos (x+y) [/tex]

Division of the two never results in tan (x+y).
Plase help me express tan (x+y) in terms of a,b!
 
Last edited:
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  • #2
Alright, I checked these two corrections and they look correct.
2ab= 2 sin (x+y)+sin 2x+sin 2y
[tex]b^{2}-a^{2}= cos 2x + cos 2y + 2 cos (x+y)[/tex]
For the part,
sin(2x)+sin(2y)
you can use the sum and difference formulas for sines and cosines you will get
(3) [tex]2 sin \left( \frac{2x+2y}{2} \right) cos \left( \frac{2x+2y}{2} \right) = 2sin(x+y)cos(x-y)[/tex]

For the part, [tex]cos(2x)+cos(2y)[/tex]
(4) [tex]2 cos \left( \frac{2x+2y}{2} \right) cos \left( \frac{2x-2y}{2} \right)=2cos(x+y)cos(x-y)[/tex]

Edited.
 
Last edited:
  • #3
konthelion said:
Since, you know that
(1) sin x + sin y= a
(2) cos x + cos y=b

Then by the sum and difference formulas, you will get
(3) [tex]2 sin \left( \frac{x+y}{2} \right) cos \left( \frac{x+y}{2} \right) = a [/tex]
(4) [tex]2 cos \left( \frac{x+y}{2} \right) cos \left( \frac{x-y}{2} \right) = b [/tex]

No, wait scratch this. This won't help.

Thanks a lot friend. I hav solved it now. Thanks once again.
 

1. What are the basic trigonometric ratios?

The basic trigonometric ratios are sine (sin), cosine (cos), and tangent (tan). They are used to relate the sides of a right triangle to its angles.

2. What is the formula for finding a trigonometric ratio?

The formula for finding a trigonometric ratio is the ratio of two sides of a right triangle. For example, the sine of an angle is equal to the opposite side divided by the hypotenuse.

3. How do you solve for a missing side using trigonometric ratios?

To solve for a missing side using trigonometric ratios, you can use the inverse function of the ratio. For example, if you are looking for the length of the opposite side, you would use the inverse sine function to find the angle, and then use the sine ratio to find the length of the opposite side.

4. Can trigonometric ratios be used for non-right triangles?

No, trigonometric ratios can only be used for right triangles. Non-right triangles require different methods for solving.

5. How do you simplify trigonometric ratios?

To simplify trigonometric ratios, you can use common factors to cancel out terms in the numerator and denominator. You can also use trigonometric identities to rewrite the ratio in a simpler form.

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