Skip to content
Exercise 3.3 · Q11

Q.Prove that cos⁡(3π4+x)−cos⁡(3π4−x)=−2 sin⁡x\cos\left(\frac{3\pi}{4}+x\right) - \cos\left(\frac{3\pi}{4}-x\right) = -\sqrt{2}\,\sin x.

Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★est
25% · 37/150 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

Using the cosine difference identity cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}, we simplify the left-hand side directly to −2sin⁡x-\sqrt{2}\sin x, proving the identity.

The key here is to avoid expanding each cosine separately with the angle-sum formula — that works but is messy. Instead, we use a sum-to-product identity, which is tailor-made for expressions of the form cos⁡P−cos⁡Q\cos P - \cos Q. This identity converts the difference of two cosines into a product of sines, and the angles simplify beautifully because the xx terms cancel in the average but survive in the half-difference.

Let’s walk through it.

  1. Recall the sum-to-product identity for cosine difference. For any two angles AA and BB:

cos⁡A−cos⁡B=−2sin⁡A+B2sin⁡A−B2.\cos A - \cos B = -2 \sin\frac{A+B}{2} \sin\frac{A-B}{2}.

This is a standard result derived from the cosine addition formulas. It’s especially useful when AA and BB are symmetric about some point — exactly our case.

  1. Identify AA and BB from the problem. Here:

A=3π4+x,B=3π4−x.A = \frac{3\pi}{4} + x, \quad B = \frac{3\pi}{4} - x.

Notice that AA and BB are symmetric: their sum is constant, and their difference depends only on xx.

  1. Compute A+B2\frac{A+B}{2} and A−B2\frac{A-B}{2}.

A+B2=(3π4+x)+(3π4−x)2=3π22=3π4.\frac{A+B}{2} = \frac{\left(\frac{3\pi}{4}+x\right) + \left(\frac{3\pi}{4}-x\right)}{2} = \frac{\frac{3\pi}{2}}{2} = \frac{3\pi}{4}.

The xx terms cancel — neat.

A−B2=(3π4+x)−(3π4−x)2=2x2=x.\frac{A-B}{2} = \frac{\left(\frac{3\pi}{4}+x\right) - \left(\frac{3\pi}{4}-x\right)}{2} = \frac{2x}{2} = x.

  1. Plug into the identity.

cos⁡(3π4+x)−cos⁡(3π4−x)=−2sin⁡(3π4)sin⁡(x).\cos\left(\frac{3\pi}{4}+x\right) - \cos\left(\frac{3\pi}{4}-x\right) = -2 \sin\left(\frac{3\pi}{4}\right) \sin(x).

  1. Evaluate sin⁡(3π/4)\sin(3\pi/4). The angle 3π/43\pi/4 is in the second quadrant, where sine is positive. Its reference angle is π/4\pi/4, so:

sin⁡3π4=sin⁡π4=22.\sin\frac{3\pi}{4} = \sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}.

  1. Substitute and simplify. −2⋅22⋅sin⁡x=−2sin⁡x.-2 \cdot \frac{\sqrt{2}}{2} \cdot \sin x = -\sqrt{2} \sin x. …

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.