Test:211 QNo:3

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cos²A+cos²B+cos²C+2cosAcosBcosC
where ABC is a triangle

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prakash ·2010-05-20 13:11:30

put A=B=C=60

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Pritish Chakraborty ·2010-05-20 22:33:17

cosÂ²A + cosÂ²B + cosÂ²C = 1 - 2cosAcosBcosC

Therefore cosÂ²A + cosÂ²B + cosÂ²C + 2cosAcosBcosC = 1

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Atul ·2010-10-18 03:20:27

this question did not appear properly in the test.............. i dint understnd the question as unnecessary terms like p and square brackets were put in the question.............

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Basic Integral Calculus Operators Symbols Matrix Cases

√ √ || {} [] () → ~ ^ x x x → → ln log lg lim lim cos sin tan cot arcsin arccos arctan

∫ ∫ ∫ ∫ ∫ ∬ ∬ ∬ ∬ ∬ ∭ ∭ ∭ ∭ ∭ ⨌ ⨌ ⨌ ⨌ ⨌

∑ ∑ ∑ ∑ ∑ ∮ ∮ ∮ ∮ ∮ ∏ ∏ ∏ ∏ ∏ ∐ ∐ ∐ ∐ ∐

+ - × ÷ ± ∓ = ∪ ∩ ≠ ∼ ≈ ∅ ∀ ∃ ∈ ∋ ∝ ≤ ≥ ≃ ≅ ≡ < > ≪ ≫ ⊂ ∉ ⊃ ⊆ ⊇ ⇒ ⇐ ⇔ ⇒ ⇐ ⇔ ⇑ ⇓ ⇕ → ← ↔ → ← ↔ ↑ ↓ ↕ ↖ ↗ ↘ ↙ ↪ ↩ ⇀ ↼ ⇁ ↽

° ∂ ∞ α β γ δ θ λ μ ν ξ π ρ σ φ ω ε ƒ κ τ υ Γ Δ Λ Σ Π Ω

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Test:211 QNo:3

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