Skip to content
Miscellaneous Exercise · Q3

Q.Prove that (cos⁡x+cos⁡y)2+(sin⁡x−sin⁡y)2=4cos⁡2x+y2(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\frac{x+y}{2}.

Yanam CbseNCERTSubjective· 3mImportance★★★★★est
45% · 67/150 Questions
✓ Free question

The identity is proved by expanding both squares, applying the Pythagorean identity to simplify the cross terms, and then using the sum-to-product formula for cosine. The final result is 4cos⁡2x+y24\cos^2\frac{x+y}{2}.

This problem is a classic exercise in trigonometric identities. The key is to resist the temptation to jump into half-angle formulas too early. Instead, start by expanding the squares — this will give you terms like cos⁡2x\cos^2 x, sin⁡2y\sin^2 y, and crucially, the cross terms 2cos⁡xcos⁡y2\cos x \cos y and −2sin⁡xsin⁡y-2\sin x \sin y. The combination of those cross terms is what collapses into a neat cosine sum formula.

Let’s work through it step by step.

  1. Expand the left-hand side

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

Expanding each square:

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

  1. Group like terms Collect the cos⁡2\cos^2 and sin⁡2\sin^2 terms together:

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

  1. Apply the Pythagorean identity For any angle, cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1. So:

=1+1+2(cos⁡xcos⁡y−sin⁡xsin⁡y)= 1 + 1 + 2(\cos x \cos y - \sin x \sin y)

=2+2(cos⁡xcos⁡y−sin⁡xsin⁡y)= 2 + 2(\cos x \cos y - \sin x \sin y)

  1. Recognize the cosine addition formula The expression inside the parentheses is exactly cos⁡(x+y)\cos(x + y):

cos⁡xcos⁡y−sin⁡xsin⁡y=cos⁡(x+y)\cos x \cos y - \sin x \sin y = \cos(x + y)

So we have:

=2+2cos⁡(x+y)= 2 + 2\cos(x + y)

  1. Factor and use the half-angle identity Factor out a 2:

=2[1+cos⁡(x+y)]= 2[1 + \cos(x + y)]

Now recall the identity: 1+cos⁡θ=2cos⁡2θ21 + \cos\theta = 2\cos^2\frac{\theta}{2}. Here θ=x+y\theta = x + y, so:

1+cos⁡(x+y)=2cos⁡2x+y21 + \cos(x + y) = 2\cos^2\frac{x + y}{2}

Substituting:

=2⋅2cos⁡2x+y2=4cos⁡2x+y2= 2 \cdot 2\cos^2\frac{x + y}{2} = 4\cos^2\frac{x + y}{2}

Watch out

A common mistake is to incorrectly expand (sin⁡x−sin⁡y)2(\sin x - \sin y)^2 as sin⁡2x−sin⁡2y\sin^2 x - \sin^2 y — that’s wrong! The correct expansion is sin⁡2x−2sin⁡xsin⁡y+sin⁡2y\sin^2 x - 2\sin x \sin y + \sin^2 y. Always write out the middle term.

Tip

If you ever see (cos⁡A+cos⁡B)2+(sin⁡A±sin⁡B)2(\cos A + \cos B)^2 + (\sin A \pm \sin B)^2, expect the cross terms to combine into cos⁡(A∓B)\cos(A \mp B). Here the minus sign in the sine term gives cos⁡(x+y)\cos(x+y); if it were (sin⁡x+sin⁡y)2(\sin x + \sin y)^2, you’d get cos⁡(x−y)\cos(x-y) instead.

✓Final answer

The identity is proved: (cos⁡x+cos⁡y)2+(sin⁡x−sin⁡y)2=4cos⁡2x+y2(\cos x + \cos y)^2 + (\sin x - \sin y)^2 = 4\cos^2\frac{x+y}{2}.

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.