Q.If acosθ−bsinθ=c, show that asinθ+bcosθ=±a2+b2−c2.
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Concept understanding — Trigonometric Identities
The six trigonometric ratios (right-triangle definition). For an acute angle θ in a right triangle, with sides labelled relative to θ as opposite, adjacent, and the hypotenuse:
Only sinθ and cosθ are truly independent — every other ratio is a quotient or reciprocal built from them, which is why rewriting a hard trig expression purely in sin,cos is such a reliable first move.
Exact values at standard angles (0∘,30∘,45∘,60∘,90∘):
θ
0∘
30∘
45∘
60∘
90∘
sinθ
0
21
21
23
1
cosθ
1
23
21
21
0
tanθ
0
31
1
3
undefined
tan90∘,sec90∘ are undefined because cos90∘=0; csc0∘,cot0∘ are undefined because sin0∘=0. Also sin30∘=cos60∘ and sin60∘=cos30∘ — an early instance of the general complementary-angle pattern sinθ=cos(90∘−θ).
What makes an equation an identity. A trigonometric identity is an equation in trigonometric ratios holding for every value of θ in its domain — not merely for some particular angle. secθ=cosθ1 is an identity (true for all θ with cosθ=0); sinθ=21 is not (true only at specific angles like 30∘ or 150∘).
The three fundamental (Pythagorean) identities, obtained from the Pythagorean theorem applied to a right triangle, divided in turn by the square of the hypotenuse, the adjacent side, and the opposite side:
cos2θ+sin2θ=1,sec2θ−tan2θ=1,csc2θ−cot2θ=1.
Here sin2θ means (sinθ)2, and similarly for the other ratios. Each identity holds wherever both sides are defined — e.g. sec2θ−tan2θ=1 says nothing at θ=90∘, where both terms are individually undefined, but this doesn't stop it from being a genuine identity for every θ where it does make sense.
Reciprocal and quotient identities (restating the ratio definitions as identities in their own right):
How identities are actually used. Proving a trig identity is an exercise in disciplined algebra layered on top of the three Pythagorean identities: combine fractions over a common denominator, factor a difference of squares (very often sec2θ−tan2θ=1 is used in its factored form (secθ−tanθ)(secθ+tanθ)=1), or use a sum/difference-of-cubes factorisation together with sin2θ+cos2θ=1. When a problem hands over two separate equations relating two unknowns to the same angle θ, a very common technique is to square both equations and add them, which usually collapses instantly via sin2θ+cos2θ=1 or sec2θ−tan2θ=1 into an equation with θ eliminated entirely.
Tip
If stuck proving an identity, try three things in order: (1) rewrite everything in sin,cos; (2) look for a Pythagorean substitution hiding in a sum/difference of squares; (3) combine all fractions into one before simplifying — simplifying term-by-term first often hides the cancellation that makes the identity work.
Square both the given relation and the target expression, and add — the cross terms cancel via sin2θ+cos2θ=1.
✓Final answer
asinθ+bcosθ=±a2+b2−c2.
Squaring acosθ−bsinθ=c and squaring the target expression E=asinθ+bcosθ, then adding the two results, eliminates θ completely because the cross terms in the two expansions are exact negatives of each other.
Step 1. Square the given relation.
(acosθ−bsinθ)2=c2
⇒a2cos2θ−2absinθcosθ+b2sin2θ=c2 … (I)
Step 2. Square the target expression E=asinθ+bcosθ.
E2=(asinθ+bcosθ)2=a2sin2θ+2absinθcosθ+b2cos2θ … (II)
Step 3. Add (I) and (II). The cross terms −2absinθcosθ and +2absinθcosθ cancel:
c2+E2=a2(cos2θ+sin2θ)+b2(sin2θ+cos2θ)=a2+b2
using sin2θ+cos2θ=1.
Step 4. Solve for E.E2=a2+b2−c2⇒E=±a2+b2−c2.
So asinθ+bcosθ=±a2+b2−c2, exactly as required — the ± is unavoidable since squaring an equation always introduces both signs.
✓Final answer
asinθ+bcosθ=±a2+b2−c2.
Sign error in the cross term when expanding one of the two squares
Dropping the ± when taking the square root at the end