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Exercise 3.3 · Q6

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

Chandigarh CbseNCERTSubjective· 3mImportance★★★★★est
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This identity is a direct application of the cosine addition formula in reverse. By recognising the left-hand side as cos⁡(A+B)\cos(A+B) with A=π4−xA = \frac{\pi}{4}-x and B=π4−yB = \frac{\pi}{4}-y, we simplify to cos⁡(π2−(x+y))\cos\left(\frac{\pi}{2} - (x+y)\right), which equals sin⁡(x+y)\sin(x+y).

The key insight here is that the expression cos⁡Acos⁡B−sin⁡Asin⁡B\cos A \cos B - \sin A \sin B is the expanded form of cos⁡(A+B)\cos(A+B). This is one of the most fundamental trigonometric identities, and once you spot it, the problem becomes almost mechanical.

Let’s walk through it step by step.

  1. Identify the pattern The left-hand side is:

cos⁡(π4−x)cos⁡(π4−y)−sin⁡(π4−x)sin⁡(π4−y)\cos\left(\frac{\pi}{4}-x\right)\cos\left(\frac{\pi}{4}-y\right) - \sin\left(\frac{\pi}{4}-x\right)\sin\left(\frac{\pi}{4}-y\right)

This matches exactly the formula cos⁡Pcos⁡Q−sin⁡Psin⁡Q=cos⁡(P+Q)\cos P \cos Q - \sin P \sin Q = \cos(P+Q).

  1. Apply the cosine addition formula Set P=π4−xP = \frac{\pi}{4} - x and Q=π4−yQ = \frac{\pi}{4} - y. Then:

LHS=cos⁡[(π4−x)+(π4−y)]\text{LHS} = \cos\left[ \left(\frac{\pi}{4}-x\right) + \left(\frac{\pi}{4}-y\right) \right]

  1. Simplify the angle Add the angles inside the cosine:

(π4−x)+(π4−y)=π2−(x+y)\left(\frac{\pi}{4}-x\right) + \left(\frac{\pi}{4}-y\right) = \frac{\pi}{2} - (x+y)

So the expression becomes:

cos⁡(π2−(x+y))\cos\left(\frac{\pi}{2} - (x+y)\right)

  1. Use the co-function identity A standard result: cos⁡(π2−θ)=sin⁡θ\cos\left(\frac{\pi}{2} - \theta\right) = \sin\theta. Therefore: cos⁡(π2−(x+y))=sin⁡(x+y)\cos\left(\frac{\pi}{2} - (x+y)\right) = \sin(x+y) …

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