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If \( 2 \cos \theta=\sqrt{x}+\frac{1}{\sqrt{x}} \) for some \( x>0 \), then which one of the following is not equal to \( \cos (5 \theta) \) (a) \( \frac{1}{2}\left(x^{5 / 2}+\frac{1}{x^{5 / 2}}\right) \) (b) \( \frac{1}{2}\left(x^{3 / 2}+\frac{1}{x^{3 / 2}}\right) \) (c) 1 (d) \( -1 \) |
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Answer» Correct option is (a) \(\frac12(x^{5/2}+\frac1{x^{5/2}})\) 2 cos θ = \(\sqrt x+\frac1{\sqrt x}\)-----(1) 4 cos2 θ = x + 1/x + 2----(2) (By squaring (2)) and 8 cos3θ = x3/2 + \(\frac1{x^{3/2}}\) + 3\(\sqrt x\) x \(\frac1{\sqrt x}\) (\(\sqrt x+\frac1{\sqrt x}\)) = x3/2 + \(\frac1{x^{3/2}}\) + 3 (\(\sqrt x+\frac1{\sqrt x}\))---(3) Now, cos 5θ = cos(4θ + θ) = cos 4θ.cosθ - sin 4θ sin θ = (2cos2θ -1) cos θ - 2 sin 2θ cos2θ sin θ = (2(2cos2θ - 1)2 - 1) cos θ - 4 sin2θ cos θ(2cos2 θ -1) = (2(4cos4θ - 4 cos2θ +1)-1) cos θ - 4(1- cos2θ)(2cos3θ - cos θ) = (8 cos4θ - 8 cos2θ + 2 - 1) cos θ - 4(-2cos5θ + 2cos3θ + cos3θ - cos θ) = 8 cos5θ - 8 cos3θ + cos θ + 8cos5θ - 12 cos3θ + 4cos θ = 16 cos5θ - 20 cos3θ + 5 cos θ = \(\frac12(8 cos^3\theta\times4cos^2\theta)-\frac52(8cos^3\theta)\) + \(\frac52\)(2cos θ) = \(\frac12((x^{3/2}+\frac1{x^{3/2}}+3(\sqrt x+\frac1{\sqrt x}))\)\((x+\frac 1x+2)\)) = \(\frac12\Big(x^{3/2}+\frac1{x^{1/2}}+3x^{3/2}+3x^{1/2}+\frac1{x^{5/2}}+\frac3{x^{1/2}}+\frac3{x^{3/2}}\) \(+2x^{3/2}+\frac2{x^{3/2}}+6x^{1/2}+\frac6{x^{1/2}}-5x^{3/2}-\frac5{x^{3/2}}\) \(-15x^{1/2}-\frac{15}{x^{1/2}}\Big)\) = \(\frac12(x^{5/2}+\frac1{x^{5/2}}+x^{3/2}(3+2-5)+\frac1{x^{3/2}}(3+2-5)\) \(+x^{1/2}(3+1+6-15+5)+\frac1{x^{1/2}}(1+3+6-15+5))\) = \(\frac12(x^{5/2}+\frac1{x^{3/2}}+x^{3/2}(5-5)+x^{1/2}(15-15)+\frac1{x^{1/2}}(15-15))\) = \(\frac12(x^{5/2}+\frac1{x^{5/2}})\) |
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