Q.If k=sin(18π)sin(185π)sin(187π), then the numerical value of k is ______.
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🔒 Start your 14-day free trial to unlock the full solution →Concept understanding — Trigonometric Identity Proof
Trigonometric Identity Proof: From Intuition to Precision
Imagine you're standing at the corner of a right triangle. The two shorter sides — one horizontal, one vertical — and the sloping hypotenuse are all connected. If you change the angle at your corner, the lengths of the sides change, but the relationship between them stays fixed. That fixed relationship is what a trigonometric identity captures.
The Core Idea
A trigonometric identity is an equation involving trigonometric functions (like sinθ, cosθ, tanθ) that is true for every angle θ where both sides are defined. It's not a conditional equation (like sinθ=0.5, which is true only for specific angles). It's an eternal truth about how these functions relate.
The most famous one is:
sin2θ+cos2θ=1
This holds for any angle θ — acute, obtuse, negative, whatever. Why? Because on the unit circle, sinθ is the y-coordinate and cosθ is the x-coordinate of a point on a circle of radius 1. The Pythagorean theorem says x2+y2=1, so sin2θ+cos2θ=1 is just the Pythagorean theorem in disguise.
Proving an Identity: The Method
When you're asked to prove a trigonometric identity, you're not solving for an angle. You're showing that the left-hand side (LHS) and right-hand side (RHS) are the same expression, just written differently.
The golden rule: Start with one side and transform it into the other, using known identities and algebraic manipulation. Never move terms across the equals sign as if solving an equation — that assumes the identity is already true, which is what you're trying to prove.
A Simple Example
Prove: tanθ⋅cosθ=sinθ
Step 1: Pick a side to start with. Usually, the more complicated side is easier to simplify. Here, the LHS looks more complex.
Step 2: Replace tanθ with cosθsinθ (a known identity).
tanθ⋅cosθ=cosθsinθ⋅cosθ
Step 3: Cancel cosθ (provided cosθ=0 — but the identity holds for all angles where both sides are defined, and at cosθ=0, tanθ is undefined anyway).
=sinθ
That's it. The LHS simplifies exactly to the RHS.
The Toolbox of Known Identities
To prove any identity, you need to know the basic building blocks:
| Identity | Formula |
|---|---|
| Pythagorean | sin2θ+cos2θ=1 |
| Quotient | tanθ=cosθsinθ, cotθ=sinθcosθ |
| Reciprocal | cscθ=sinθ1, secθ=cosθ1, cotθ=tanθ1 |
| Even-Odd | sin(−θ)=−sinθ, cos(−θ)=cosθ |
A common mistake is to treat sin2θ as (sinθ)2 — which it is — but then incorrectly think sin2θ+cos2θ=1 means sinθ+cosθ=1. It does not. The square applies to the whole sine value, not to the angle.
A Slightly Harder Proof
Prove: cosθ1−cos2θ=sinθtanθ
Start with LHS: cosθ1−cos2θ
From the Pythagorean identity, 1−cos2θ=sin2θ. So:
cosθsin2θ=sinθ⋅cosθsinθ=sinθtanθ
That's the RHS. Done. …
Concept: Product-to-sum identities and symmetry in sine products
We need to evaluate k=sin(18π)sin(185π)sin(187π).
First, observe that 187π=2π−182π=2π−9π, so sin(187π)=cos(9π).
Also note that 18π=10°, 185π=50°, and 187π=70°. These angles are 10°, 50°, and 70°. …
Writing the angles as 10∘,60∘−10∘,60∘+10∘ and using sinθsin(60∘−θ)sin(60∘+θ)=41sin3θ gives k=81.
Convert to degrees: 18π=10∘, 185π=50∘, 187π=70∘, so
k=sin10∘sin50∘sin70∘ …
- AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQQ.The value of sin(45∘+θ)−cos(45∘−θ) is: (A) 2cosθ (B) 2sinθ (C) 1 (D) 0
›Reveal solutionSolution
cos(45∘−θ)=sin(45∘+θ), so their difference is 0.
Use cosα=sin(90∘−α) with α=45∘−θ:
cos(45∘−θ)=sin(90∘−(45∘−θ))=sin(45∘+θ).
…
- AHSEC Higher Secondary (HS) 1st Year Examination 2026Set ANNUAL1 markMCQQ.In △ABC, tanA+tanB+tanC=0. Then cotA⋅cotB⋅cotC=? (A) 6 (B) 1 (C) ∞ (D) −1
›Reveal solutionSolution
Triangle identity gives tanAtanBtanC=0; hence cotAcotBcotC=tanAtanBtanC1→∞.
For a triangle, A+B+C=π, which gives the identity
tanA+tanB+tanC=tanA⋅tanB⋅tanC.
We are told tanA+tanB+tanC=0, so tanA⋅tanB⋅tanC=0.
…
- AHSEC Higher Secondary (HS) 1st Year Examination 2025Set ANNUAL1 markMCQQ.The value of sin127πcos4π−cos127πsin4π=?(a) 21(b) 2(c) 23(d) None
›Reveal solutionSolution
The expression is sin(127π−4π)=sin3π=23.
Recall the identity sinAcosB−cosAsinB=sin(A−B).
Here A=127π and B=4π=123π, so …
- AHSEC Higher Secondary (HS) 1st Year Examination 2025Set ANNUAL1 markMCQQ.The value of tan1213π is:(a) 3−2(b) 2−3(c) 2+3(d) None
›Reveal solutionSolution
tan1213π=tan15∘=2−3.
Because tan has period π,
tan1213π=tan(π+12π)=tan12π.
Now 12π=15∘ and …
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