[ASK} Prove (cos2x+cos2y)/(sin2x−sin2y)=1/tan(x−y)

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Thank you!In summary, the conversation discusses proving the identity \frac{cos2x+cos2y}{sin2x-sin2y}=\frac1{tan(x-y)} by manipulating the LHS using trigonometric identities. The final result is cot(x-y) or \frac1{tan(x-y)}.
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Monoxdifly
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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.
 
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  • #2
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.

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*}
 
  • #3
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*}

Ah, let's see...
\(\displaystyle \frac{cos2x+cos2y}{sin2x-sin2y}\)=\(\displaystyle \frac{2cos\frac{2x+2y}{2}cos\frac{2x-2y}{2}}{2sin\frac{2x-2y}{2}cos\frac{2x+2y}{2}}\)=\(\displaystyle \frac{cos(x-y)}{sin(x-y)}\)= cot(x - y) = \(\displaystyle \frac1{tan(x-y)}\)
Wew. Just 4 steps.
 
Last edited:

FAQ: [ASK} Prove (cos2x+cos2y)/(sin2x−sin2y)=1/tan(x−y)

What is the given equation?

The given equation is (cos2x+cos2y)/(sin2x−sin2y)=1/tan(x−y).

How do you prove the given equation?

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.

What are the steps to proving the given equation?

The steps to proving the given equation are:

  1. Use the double angle formula for cosine to expand cos2x and cos2y.
  2. Use the double angle formula for sine to expand sin2x and sin2y.
  3. Substitute the expanded forms into the original equation.
  4. Use the quotient identity for tangent to simplify the right side of the equation.
  5. Combine like terms and use algebraic manipulations to simplify the left side of the equation.
  6. Show that both sides of the equation are equal.

Can you provide an example of how to prove the given equation?

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.

Why is the given equation important in mathematics?

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.

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