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Exercise 3.1 · Q3

Q.If acos⁡θ−bsin⁡θ=ca\cos\theta - b\sin\theta = c, show that asin⁡θ+bcos⁡θ=±a2+b2−c2a\sin\theta + b\cos\theta = \pm\sqrt{a^2+b^2-c^2}.

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✓ Free question

Squaring acos⁡θ−bsin⁡θ=ca\cos\theta-b\sin\theta=c and squaring the target expression E=asin⁡θ+bcos⁡θE=a\sin\theta+b\cos\theta, then adding the two results, eliminates θ\theta completely because the cross terms in the two expansions are exact negatives of each other.

Step 1. Square the given relation.

(acos⁡θ−bsin⁡θ)2=c2(a\cos\theta-b\sin\theta)^2=c^2

⇒a2cos⁡2θ−2absin⁡θcos⁡θ+b2sin⁡2θ=c2\Rightarrow a^2\cos^2\theta-2ab\sin\theta\cos\theta+b^2\sin^2\theta=c^2 … (I)

Step 2. Square the target expression E=asin⁡θ+bcos⁡θE=a\sin\theta+b\cos\theta.

E2=(asin⁡θ+bcos⁡θ)2=a2sin⁡2θ+2absin⁡θcos⁡θ+b2cos⁡2θE^2=(a\sin\theta+b\cos\theta)^2=a^2\sin^2\theta+2ab\sin\theta\cos\theta+b^2\cos^2\theta … (II)

Step 3. Add (I) and (II). The cross terms −2absin⁡θcos⁡θ-2ab\sin\theta\cos\theta and +2absin⁡θcos⁡θ+2ab\sin\theta\cos\theta cancel:

c2+E2=a2(cos⁡2θ+sin⁡2θ)+b2(sin⁡2θ+cos⁡2θ)=a2+b2c^2+E^2 = a^2(\cos^2\theta+\sin^2\theta) + b^2(\sin^2\theta+\cos^2\theta) = a^2+b^2

using sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1.

Step 4. Solve for EE. E2=a2+b2−c2⇒E=±a2+b2−c2E^2=a^2+b^2-c^2 \Rightarrow E=\pm\sqrt{a^2+b^2-c^2}.

So asin⁡θ+bcos⁡θ=±a2+b2−c2a\sin\theta+b\cos\theta=\pm\sqrt{a^2+b^2-c^2}, exactly as required — the ±\pm is unavoidable since squaring an equation always introduces both signs.

✓Final answer

asin⁡θ+bcos⁡θ=±a2+b2−c2a\sin\theta+b\cos\theta=\pm\sqrt{a^2+b^2-c^2}.

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