Question and Answers Forum
TrigonometryQuestion and Answers: Page 4
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Question and Answers Forum |
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TrigonometryQuestion and Answers: Page 4 |
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| ⋐ π |
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| determinant () |
| cos (((2π)/(21)))cos (((4π)/(21)))cos (((8π)/(21)))cos (((10π)/(22)))cos (((16π)/(21)))cos (((20π)/(21)))=? |
| (1/(cos x−cos 3x)) + (1/(cos x−cos 5x)) + (1/(cos x−cos 7x)) + (1/(cos x−cos 11x))=? |
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| 3(sinθ − cosθ)^4 + 6(sinθ + cosθ)^2 + 4(sin^6 θ + cos^6 θ) = ? |
| Relating to question 207407 x^3 −12x^2 +27x−17=0 Let x=t+4 t^3 −21t−37=0 The Trigonometric Solution gives these: x_1 =4−2(√7)cos ((π+2sin^(−1) ((37(√7))/(98)))/6) x_2 =4−2(√7)sin ((sin^(−1) ((37(√7))/(98)))/3) x_3 =4+2(√7)sin ((π+sin^(−1) ((37(√7))/(98)))/3) Prove these identities: x_1 =2−((1+2sin (π/(18)))/(2cos (π/9))) x_2 =2+((1+2cos (π/9))/(2cos ((2π)/9))) x_3 =((1+2(√3)sin ((2π)/9))/(2sin (π/(18)))) |
| f(x)=[cos2x+cos3x][cos4x+cos6x][[cosx+cos5x] evaluar f(((2π)/(13))) |
| Construct an angle whose sine is (3/(2 + (√5))) . |
| If sinθ = ((m^2 + 2mn)/(m^2 + 2mn + 2n^2 )) then prove that tanθ = ((m^2 + 2mn)/(2mn + 2n^2 )) . |
| If tanθ = ((2x(x + 1))/(2x + 1)) then find sinθ and cosθ. |
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| If A = sin^4 θ + cos^4 θ then select the correct option: i) 0 < A < (1/2) ii) 1 < A < (3/2) iii) (1/2) ≤ A ≤ 1 iv) (3/2) ≤ A ≤ 2 |
| If asin^2 θ + bcos^2 θ = c, bsin^2 φ + acos^2 φ = d and atanθ = btanφ then prove that (1/a) + (1/b) = (1/c) + (1/d) . |
| If asinθ = bcosθ = ((2ctanθ)/(1 − tan^2 θ)) then prove that (a^2 − b^2 )^2 = 4c^2 (a^2 + b^2 ). |
| If tan^2 θ = 1 − x^2 then prove that secθ + tan^3 θcosecθ = (√((2 − x^2 )^3 )) . |
| If tanpθ = ptanθ then prove that ((sin^2 pθ)/(sin^2 θ)) = (p^2 /(1 + (p^2 − 1)sin^2 θ)) . |
| sin(π/7) × sin((2π)/7) × sin((3π)/7) = ? |
| Prove that 2^(sin^2 θ) + 2^(cos^2 θ) ≥ 2(√2). |