Question 44 of 50intermediate🔧 ApplyLong Answer4 marks

Prove that:

(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2=4sin⁡2x−y2(\cos x - \cos y)^2 + (\sin x - \sin y)^2 = 4 \sin^2 \frac{x - y}{2}
Correct Answer

We begin with the Left Hand Side (LHS) of the equation:

(cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2(\cos x - \cos y)^2 + (\sin x - \sin y)^2

To simplify this expression, we use the transformation formulae for the difference of cosine and sine functions.

From the provided formulae, we substitute the values for the difference of angles. We use the identity:

cos⁡x−cos⁡y=−2sin⁡x+y2sin⁡x−y2\cos x - \cos y = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}

and

sin⁡x−sin⁡y=2cos⁡x+y2sin⁡x−y2\sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}

Substituting these into the LHS, we get:

[−2sin⁡x+y2sin⁡x−y2]2+[2cos⁡x+y2sin⁡x−y2]2\left[-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]^2 + \left[2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right]^2

Next, we expand the squared terms. This yields:

4sin⁡2x+y2sin⁡2x−y2+4cos⁡2x+y2sin⁡2x−y24 \sin^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2} + 4 \cos^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2}

We can factor out the common term 4sin⁡2x−y24 \sin^2 \frac{x-y}{2} from the expression.

The expression becomes:

4sin⁡2x−y2[sin⁡2x+y2+cos⁡2x+y2]4 \sin^2 \frac{x-y}{2} \left[\sin^2 \frac{x+y}{2} + \cos^2 \frac{x+y}{2}\right]

Using the fundamental trigonometric identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, the term inside the bracket becomes 11.

Thus, the expression simplifies to:

4sin⁡2x−y24 \sin^2 \frac{x-y}{2}

which is equal to the Right Hand Side (RHS). Hence, LHS = RHS, and the identity is proved.

Exercise: MISCELLANEOUS EXERCISE ON CHAPTER 3 | Q: 4 | (Page 29)
For More Understanding

Explanation

The solution utilizes the sum-to-product transformation formulae provided in the context (specifically formulae 20(ii) and 20(iv)) to expand the terms (cos⁡x−cos⁡y)(\cos x - \cos y) and (sin⁡x−sin⁡y)(\sin x - \sin y).

After substitution and squaring, algebraic simplification involving the Pythagorean identity (sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1) leads to the final result.

Solution Steps

  1. Step 1: Identify LHS as (cos⁡x−cos⁡y)2+(sin⁡x−sin⁡y)2(\cos x - \cos y)^2 + (\sin x - \sin y)^2.

  2. Step 2: Apply transformation formulae: cos⁡x−cos⁡y=−2sin⁡x+y2sin⁡x−y2\cos x - \cos y = -2 \sin \frac{x+y}{2} \sin \frac{x-y}{2} and sin⁡x−sin⁡y=2cos⁡x+y2sin⁡x−y2\sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}.

  3. Step 3: Substitute into LHS to get [−2sin⁡x+y2sin⁡x−y2]2+[2cos⁡x+y2sin⁡x−y2]2\left[-2 \sin \frac{x+y}{2} \sin \frac{x-y}{2}\right]^2 + \left[2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\right]^2.

  4. Step 4: Expand squares to get 4sin⁡2x+y2sin⁡2x−y2+4cos⁡2x+y2sin⁡2x−y24 \sin^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2} + 4 \cos^2 \frac{x+y}{2} \sin^2 \frac{x-y}{2}.

  5. Step 5: Factor out 4sin⁡2x−y24 \sin^2 \frac{x-y}{2} to get 4sin⁡2x−y2[sin⁡2x+y2+cos⁡2x+y2]4 \sin^2 \frac{x-y}{2} \left[\sin^2 \frac{x+y}{2} + \cos^2 \frac{x+y}{2}\right].

  6. Step 6: Use identity sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1 to simplify the bracket to 11.

  7. Step 7: Conclude LHS = 4sin⁡2x−y24 \sin^2 \frac{x-y}{2} = RHS.

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