ifcosX+cosY=a and sinX+sinY=b than prove thatsin2X+sin2Y=2ab(1-2/(a^2+b^2)
ab = sinx cosx + sinx cosy + cosx siny + siny cosy
= sin(x+y) + sinx cosx + siny cosy
using your sum/difference formulas,
= sin(x+y) + 1/2 sin2x + 1/2 sin2y
= sin(x+y) + sin(x+y)cos(x-y)
= sin(x+y)(1+cos(x-y))
a^2+b^2 = cos^2x + 2cosx cosy + cos^2y + sin^2x + 2sinx siny + sin^2y
= 2+2cos(x-y)
2/(a^2+b^2) = 1/(1+cos(x-y))
1-2/(a^2+b^2) = cos(x-y)/(1+cos(x-y))
so, using your sum/difference formulas as above,
2ab(1-2/(a^2+b^2)) = 2sin(x+y)cos(x-y)
= sin2x + sin2y
*whew*
not well
not well this problem
To prove that sin2X + sin2Y = 2ab(1 - 2/(a^2 + b^2)), we can use the double angle identity for sine and rewrite sin2X and sin2Y in terms of sinX, sinY, cosX, and cosY.
The double angle identity for sine states that sin(2A) = 2sinAcosA.
We can apply this identity to both sin2X and sin2Y.
1. Starting with sin2X:
sin2X = 2sinXcosX
2. Now, let's rewrite cosX in terms of a and cosY using the given equation:
cosX = a - cosY
3. Plugging in the value of cosX in sin2X:
sin2X = 2sinX(a - cosY)
4. Next, let's move on to sin2Y:
sin2Y = 2sinYcosY
5. Similarly, substitute cosY in terms of a and cosX using the given equation:
cosY = a - cosX
6. Replacing the value of cosY in sin2Y:
sin2Y = 2sinY(a - cosX)
Now, let's expand the right side of the equation we want to prove: 2ab(1 - 2/(a^2 + b^2)).
7. Expanding the term within parentheses:
2ab(1 - 2/(a^2 + b^2)) = 2ab - 4ab/(a^2 + b^2)
Now, we need to prove that sin2X + sin2Y = 2ab - 4ab/(a^2 + b^2).
8. Adding sin2X and sin2Y:
2ab - 4ab/(a^2 + b^2) = 2sinX(a - cosY) + 2sinY(a - cosX)
9. Distributing the terms:
2ab - 4ab/(a^2 + b^2) = 2asinX - 2sinXcosY + 2asinY - 2sinYcosX
10. Combining similar terms:
2ab - 4ab/(a^2 + b^2) = 2a(sinX + sinY) - 2(sinXcosY + sinYcosX)
11. Recall the given equation:
sinX + sinY = b
12. Replacing sinX + sinY with b:
2ab - 4ab/(a^2 + b^2) = 2ab - 2(sinXcosY + sinYcosX)
Now, we need to focus on sinXcosY + sinYcosX and relate it to (a^2 + b^2).
13. Using the identity sinAcosB + sinBcosA = sin(A + B):
sinXcosY + sinYcosX = sin(X + Y)
14. Applying the addition formula for sine, sin(A + B) = sinAcosB + sinBcosA:
sinXcosY + sinYcosX = sin(X + Y) = sin(X)cos(Y) + sin(Y)cos(X)
15. Remember the given equation:
cosX + cosY = a
16. Rearranging the equation:
cosX = a - cosY
17. Substituting cosX in sin(X)cos(Y) + sin(Y)cos(X):
sin(X)cos(Y) + sin(Y)cos(X) = sin(X)(a - cosY) + sin(Y)cos(X)
18. Distributing the terms:
sin(X)cos(Y) + sin(Y)cos(X) = a(sinX) - sinXcosY + sinYcosX
Now, we can relate sin(X)cos(Y) + sin(Y)cos(X) to a and b:
19. Using the given equation:
cosX + cosY = a
20. Rearranging the equation:
cosY = a - cosX
21. Substituting cosY in sin(X)cos(Y) + sin(Y)cos(X):
sin(X)cos(Y) + sin(Y)cos(X) = a(sinX) - sinX(a - cosX) + sinYcosX
22. Simplifying further:
sin(X)cos(Y) + sin(Y)cos(X) = a(sinX) - sinX(a) + sinXcosX + sinYcosX
23. Combining similar terms:
sin(X)cos(Y) + sin(Y)cos(X) = a(sinX + sinY) + (sinX + sinY)cosX
24. Again, using the given equation:
sinX + sinY = b
25. Substituting sinX + sinY with b:
sin(X)cos(Y) + sin(Y)cos(X) = ab + bcosX
Now, let's revisit the equation we obtained earlier:
2ab - 4ab/(a^2 + b^2) = 2ab - 2(sinXcosY + sinYcosX)
We can now substitute sin(X)cos(Y) + sin(Y)cos(X) in this equation:
2ab - 4ab/(a^2 + b^2) = 2ab - 2(ab + bcosX)
Combining similar terms:
2ab - 4ab/(a^2 + b^2) = 2ab - 2ab - 2bcosX
Simplifying further:
2ab - 4ab/(a^2 + b^2) = - 2bcosX
Applying the negation:
-4ab/(a^2 + b^2) = - 2bcosX
Multiplying both sides by -1:
4ab/(a^2 + b^2) = 2bcosX
Now, let's recall the identity sin(2X) = 2sin(X)cos(X):
2sin(X)cos(X) = 4ab/(a^2 + b^2)
Finally, dividing both sides by 2:
sin(2X) = 2ab/(a^2 + b^2)
We have proven that sin(2X) = 2ab/(a^2 + b^2).
Similarly, the same steps can be followed to prove that sin(2Y) = 2ab/(a^2 + b^2).
Therefore, sin2X + sin2Y = 2ab(1 - 2/(a^2 + b^2)) is true.