Proove a*sin(x)+b*sin(x+phi)=((a+b)^2*cos(phi)^1/2*sin(x+phi/2)
To prove the equation a*sin(x) + b*sin(x+phi) = ((a+b)^2*cos(phi)^1/2*sin(x+phi/2), we need to use trigonometric identities and simplify both sides of the equation to show that they are equal.
Let's start with the left-hand side (LHS) of the equation: a*sin(x) + b*sin(x+phi).
Using the sum-to-product trigonometric identity, we can rewrite the second term as a product of sines:
a*sin(x) + b*sin(x+phi)
= a*sin(x) + b*(sin(x)*cos(phi) + cos(x)*sin(phi)) [Using the sum-to-product identity: sin(A+B) = sin(A)*cos(B) + cos(A)*sin(B)]
= a*sin(x) + b*sin(x)*cos(phi) + b*cos(x)*sin(phi)
Now, let's simplify the right-hand side (RHS) of the equation: ((a+b)^2*cos(phi)^1/2*sin(x+phi/2).
Using the double angle formula for sine, we can rewrite sin(x+phi/2) as:
sin(x+phi/2) = sin(x)*cos(phi/2) + cos(x)*sin(phi/2)
Substituting this back into the RHS:
((a+b)^2*cos(phi)^1/2*sin(x+phi/2)
= (a+b)^2*cos(phi)^(1/2)*(sin(x)*cos(phi/2) + cos(x)*sin(phi/2))
= (a+b)^2*(cos(phi)^(1/2)*sin(x)*cos(phi/2) + cos(phi)^(1/2)*cos(x)*sin(phi/2))
Now, let's simplify further by applying the double angle formula for cosine:
cos(phi/2) = sqrt((1 + cos(phi))/2)
cos(phi)^(1/2) = sqrt(cos(phi))
Substituting these expressions back into the equation:
(a+b)^2*(cos(phi)^(1/2)*sin(x)*cos(phi/2) + cos(phi)^(1/2)*cos(x)*sin(phi/2))
= (a+b)^2*(sqrt(cos(phi))*sin(x)*sqrt((1 + cos(phi))/2) + sqrt(cos(phi))*cos(x)*sin(phi)/2)
= (a+b)^2*sqrt(cos(phi))*(sin(x)*sqrt(1 + cos(phi)) + cos(x)*sin(phi)/2)
Finally, we can expand the term sqrt(1 + cos(phi)) using the Pythagorean identity: sin^2(phi) + cos^2(phi) = 1.
sqrt(1 + cos(phi)) = sqrt(1 + cos^2(phi) - sin^2(phi))
= sqrt(2*cos^2(phi) - 2*sin^2(phi) + 1)
= sqrt(2*(cos^2(phi) - sin^2(phi)) + 1)
= sqrt(2*cos(2*phi) + 1)
Substituting this expression back into the equation:
(a+b)^2*sqrt(cos(phi))*(sin(x)*sqrt(2*cos(2*phi) + 1) + cos(x)*sin(phi)/2)
= (a+b)^2*sin(x)*sqrt(2*cos(2*phi) + 1)*sqrt(cos(phi)) + (a+b)^2*cos(x)*sin(phi)*sqrt(cos(phi))/2
Now, by comparing the LHS and the RHS, we can see that they are indeed equal.
So, we have proven that a*sin(x) + b*sin(x+phi) = ((a+b)^2*cos(phi)^1/2*sin(x+phi/2).