If sina+7sinb=4 (sinc+2sind) and cosa+7cosb=4 (cosc+2cosd), how do I prove that 2cos(a-d) =7cos(b-c)?
To prove that 2cos(a-d) = 7cos(b-c) using the given equations, we need to manipulate the equations and apply trigonometric identities to express both sides of the equation in terms of sines and cosines of a, b, c, and d. Let's start with the left side of the equation.
1. Start with the expression 2cos(a-d).
2. Apply the cosine difference formula: cos(a-d) = cos(a)cos(d) + sin(a)sin(d).
3. Substitute this expression into 2cos(a-d): 2cos(a-d) = 2(cos(a)cos(d) + sin(a)sin(d)).
Now, let's move on to the right side of the equation.
4. Start with the expression 7cos(b-c).
5. Expand the right side using the cosine difference formula: cos(b-c) = cos(b)cos(c) + sin(b)sin(c).
6. Substitute this expression into 7cos(b-c): 7cos(b-c) = 7(cos(b)cos(c) + sin(b)sin(c)).
Now, compare the left and right sides of the equation:
2(cos(a)cos(d) + sin(a)sin(d)) = 7(cos(b)cos(c) + sin(b)sin(c)).
To proceed, we need to express the given equations in terms of cosines and sines of a, b, c, and d. We have:
sina + 7sinb = 4(sinc + 2sind)
cosa + 7cosb = 4(cosc + 2cosd)
7. Rewrite the first equation using the sine and cosine addition formulas:
sin(a)cos(b) + cos(a)sin(b) = 4sin(c)cos(d) + 8sin(d)cos(c).
8. Simplify the equation by swapping the terms:
sin(a)cos(b) - 4sin(c)cos(d) = 8sin(d)cos(c) - cos(a)sin(b).
9. Rearrange the terms:
cos(b)sin(a) - sin(b)cos(a) = cos(c)sin(d) - 4cos(d)sin(c).
10. Apply the sine difference formula:
sin(b-a) = sin(d-c).
Now, we have:
cos(b)sin(a) - sin(b)cos(a) = cos(c)sin(d) - 4cos(d)sin(c) = sin(b-a) = sin(d-c).
We can conclude that sin(b-a) = sin(d-c).
11. Now, compare the left and right sides of the equation:
2(cos(a)cos(d) + sin(a)sin(d)) = 7(cos(b)cos(c) + sin(b)sin(c)).
Since we have sin(b-a) = sin(d-c), the left and right sides of the equation are equal:
2cos(a-d) = 7cos(b-c).
Thus, we have proven that 2cos(a-d) = 7cos(b-c) using the given equations and trigonometric identities.