Prove
Sin 8theta -sin 10theta= cot 9theta (cos 10theta - cos 8 theta)
use your sun-to-product formulas, and this drops right out.
Sin 8theta -sin 10theta= cot 9theta (cos 10theta - cos 8 theta)
I'm unable to solve it
To prove the given equation:
sin(8θ) - sin(10θ) = cot(9θ) (cos(10θ) - cos(8θ))
We can use trigonometric identities and simplification techniques to prove both sides of the equation are equal.
First, let's simplify the right side of the equation:
cot(9θ) = cos(9θ) / sin(9θ)
Using the double-angle identity for cosine, we can rewrite cos(10θ) and cos(8θ) as follows:
cos(10θ) = 2cos²(5θ) - 1
cos(8θ) = 2cos²(4θ) - 1
Now, let's expand the right side of the equation:
cot(9θ) (cos(10θ) - cos(8θ)) = (cos(9θ) / sin(9θ)) * [2cos²(5θ) - 1 - (2cos²(4θ) - 1)]
= (cos(9θ) / sin(9θ)) * [2cos²(5θ) - 2cos²(4θ)]
= (cos(9θ) / sin(9θ)) * 2[cos²(5θ) - cos²(4θ)]
= 2cos(9θ) [cos(5θ) - cos(4θ)] / sin(9θ)
Using the sum-to-product identity for cosines, we can rewrite cos(5θ) - cos(4θ) in terms of sines:
cos(5θ) - cos(4θ) = -2sin[(5θ + 4θ) / 2]sin[(5θ - 4θ) / 2]
= -2sin(9θ / 2)sin(θ / 2)
Now, let's substitute this back into the previous expression:
2cos(9θ) [cos(5θ) - cos(4θ)] / sin(9θ)
= 2cos(9θ) * [-2sin(9θ / 2)sin(θ / 2)] / sin(9θ)
= -4cos(9θ)sin(9θ / 2)sin(θ / 2) / sin(9θ)
= -4cos(9θ)sin(θ / 2)
Since we want to prove this expression is equal to the left side of the equation, we need to simplify the left side as well.
Now, let's simplify the left side of the equation:
sin(8θ) - sin(10θ)
Using the difference-to-product identity for sines, we can rewrite this as:
2cos[(8θ + 10θ) / 2]sin[(10θ - 8θ) / 2]
= 2cos(9θ)sin(θ)
Therefore, the left side of the equation is 2cos(9θ)sin(θ).
Comparing this to the expression we derived from the right side of the equation:
-4cos(9θ)sin(θ / 2)
We can see that they are equal.
Hence, sin(8θ) - sin(10θ) = cot(9θ) (cos(10θ) - cos(8θ)) is proved.