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Question Number 18830 by Joel577 last updated on 30/Jul/17
If sin 2x = a find the simplest expression for cos x
$$\mathrm{If}\:\mathrm{sin}\:\mathrm{2}{x}\:=\:{a} \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{simplest}\:\mathrm{expression}\:\mathrm{for}\:\mathrm{cos}\:{x} \\ $$
Commented by mrW1 last updated on 30/Jul/17
cos 2x=±(√(1−a^2 )) 2cos^2 x−1=±(√(1−a^2 )) cos^2 x=((1±(√(1−a^2 )))/2) cos x=±((√(1±(√(1−a^2 ))))/( (√2)))
$$\mathrm{cos}\:\mathrm{2x}=\pm\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{2cos}^{\mathrm{2}} \:\mathrm{x}−\mathrm{1}=\pm\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} } \\ $$$$\mathrm{cos}^{\mathrm{2}} \:\mathrm{x}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{x}=\pm\frac{\sqrt{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{a}^{\mathrm{2}} }}}{\:\sqrt{\mathrm{2}}} \\ $$
Commented by Joel577 last updated on 31/Jul/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by 433 last updated on 30/Jul/17
2sin x cos x = a 4 sin^2 x cos^2 x=a^2 4(1−cos^2 x)cos^2 x=a^2 cos^2 x=b 4b−4b^2 =a^2 4b^2 −4b+a^2 =0 Δ=16−16a^2 =16(1−a^2 ) b=((4±(√(16×(1−a^2 ))))/8)=((1±(√(1−a^2 )))/2) =((1±(√(1−sin^2 (2x))))/2)=((1±cos 2x)/2) cos 2x =2cos^2 −1 b=((1+cos 2x)/2)=((1+(√(1−a^2 )))/2)
$$ \\ $$$$ \\ $$$$\mathrm{2sin}\:{x}\:\mathrm{cos}\:{x}\:=\:{a} \\ $$$$\mathrm{4}\:\mathrm{sin}\:^{\mathrm{2}} {x}\:\mathrm{cos}\:^{\mathrm{2}} {x}={a}^{\mathrm{2}} \\ $$$$\mathrm{4}\left(\mathrm{1}−\mathrm{cos}\:^{\mathrm{2}} {x}\right)\mathrm{cos}\:^{\mathrm{2}} {x}={a}^{\mathrm{2}} \\ $$$$\mathrm{cos}^{\mathrm{2}} \:{x}={b} \\ $$$$\mathrm{4}{b}−\mathrm{4}{b}^{\mathrm{2}} ={a}^{\mathrm{2}} \\ $$$$\mathrm{4}{b}^{\mathrm{2}} −\mathrm{4}{b}+{a}^{\mathrm{2}} =\mathrm{0} \\ $$$$\Delta=\mathrm{16}−\mathrm{16}{a}^{\mathrm{2}} =\mathrm{16}\left(\mathrm{1}−{a}^{\mathrm{2}} \right) \\ $$$${b}=\frac{\mathrm{4}\pm\sqrt{\mathrm{16}×\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}}{\mathrm{8}}=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}\pm\sqrt{\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \left(\mathrm{2}{x}\right)}}{\mathrm{2}}=\frac{\mathrm{1}\pm\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}} \\ $$$$\mathrm{cos}\:\mathrm{2}{x}\:=\mathrm{2cos}\:^{\mathrm{2}} −\mathrm{1} \\ $$$${b}=\frac{\mathrm{1}+\mathrm{cos}\:\mathrm{2}{x}}{\mathrm{2}}=\frac{\mathrm{1}+\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }}{\mathrm{2}} \\ $$
Commented by Joel577 last updated on 31/Jul/17
thank you very much
$${thank}\:{you}\:{very}\:{much} \\ $$

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