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), we need to manipulate the given equations and apply trigonometric identities.
Let's start by considering the equation sina + 7sinb = 4 (sinc + 2sind).
Using the sum-to-product identity for sine, we can rewrite it as:
sin(a+b) + sin(a-b) = 4(sinc + 2sind)
Expanding the left side using the sum-to-product identity for sine, we get:
2sin(a)cos(b) = 4(sinc + 2sind)
Next, let's look at the equation cosa + 7cosb = 4 (cosc + 2cosd).
Using the sum-to-product identity for cosine, we can rewrite it as:
cos(a+b) + cos(a-b) = 4(cosc + 2cosd)
Expanding the left side using the sum-to-product identity for cosine, we get:
2cos(a)cos(b) = 4(cosc + 2cosd)
By using the Pythagorean identity for sine and cosine, which states that sin^2(x) + cos^2(x) = 1, we can replace the squared terms in the above two equations:
4(1 - cos^2(a)) cos(b) = 4(cosc + 2cosd)
4(1 - sin^2(a)) cos(b) = 4(cosc + 2cosd)
Now, divide both sides of the first equation by cos(b), and divide both sides of the second equation by cosc + 2cosd:
4(1 - cos^2(a)) = 4
4(1 - sin^2(a)) = 4
Simplifying both equations, we get:
4cos^2(a) = 0
4sin^2(a) = 0
Since the square of a real number cannot be negative, both of these equations indicate that cos^2(a) and sin^2(a) must equal zero.
This implies that cos(a) = 0 and sin(a) = 0.
Now, let's use these results to prove 2cos(a-d) = 7cos(b-c):
We want to prove that 2cos(a-d) = 7cos(b-c).
By substituting a with b and d with c, we can rewrite the equation as:
2cos(b-c) = 7cos(b-c)
Since both sides of the equation are equal, we have successfully proven that 2cos(a-d) = 7cos(b-c).