Monoxdifly said:Prove that \(\displaystyle \frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)}\). Can someone provide me some hints? I tried to manipulate the right-hand expression but got back to square one.
castor28 said:Hi Monoxdifly,
You could start with the LHS and the identities:
\begin{align*}
\cos a + \cos b &= 2\cos\frac{a+b}{2}\cos\frac{a-b}{2}\\
\sin a - \sin b &= 2\sin\frac{a-b}{2}\cos\frac{a+b}{2}
\end{align*}
The given equation is (cos2x+cos2y)/(sin2x−sin2y)=1/tan(x−y).
To prove the given equation, we need to use the trigonometric identities for cosine and sine, along with the quotient identity for tangent. We can also use algebraic manipulations to simplify the equation and show that both sides are equal.
The steps to proving the given equation are:
Yes, for example, we can start with the left side of the equation and use the double angle formulas for cosine and sine:
(cos2x+cos2y)/(sin2x−sin2y) = (cos^2x-sin^2x + cos^2y-sin^2y)/(2sinxcosx-2sinycosy)
Next, we can use the quotient identity for tangent to simplify the right side:
(cos^2x-sin^2x + cos^2y-sin^2y)/(2sinxcosx-2sinycosy) = (cos^2x-sin^2x + cos^2y-sin^2y)/(sin(x-y)cos(x+y))
Then, we can use algebraic manipulations to simplify the left side:
(cos^2x-sin^2x + cos^2y-sin^2y)/(sin(x-y)cos(x+y)) = (cos^2x+cos^2y)/(sin(x-y)cos(x+y))
Finally, we can use the double angle formula for cosine again and show that both sides are equal:
(cos^2x+cos^2y)/(sin(x-y)cos(x+y)) = (cos2x+cos2y)/(sin2x-sin2y) = 1/tan(x-y)
Therefore, we have proven the given equation.
The given equation is important in mathematics because it is a fundamental trigonometric identity that is used in many applications, such as solving equations in physics and engineering, and in geometric proofs. It also helps to understand the relationships between the trigonometric functions and how they can be manipulated to simplify equations.