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Question Number 11456 by Joel576 last updated on 26/Mar/17

sin x − cos x = a, (π/4) ≤ x ≤ (π/2) Which statement is correct? (1) sin^2 x − cos^2 x = −(1/2)(√(3 + 2a^2 − a^4 )) (2) sin^4 x + cos^4 x = (1/8)(−3a^4 + 6a^2 + 5) (3) sin 2x = ((1 − a^2 )/2) (4) tan^2 x + cot^2 x = ((−3a^4 − 2a^2 + 8a + 13 )/(2(a^4 + 2a^2 + 1)))

$$\mathrm{sin}\:{x}\:−\:\mathrm{cos}\:{x}\:=\:{a},\:\:\frac{\pi}{\mathrm{4}}\:\leqslant\:{x}\:\leqslant\:\frac{\pi}{\mathrm{2}} \\ $$$$\mathrm{Which}\:\mathrm{statement}\:\mathrm{is}\:\mathrm{correct}? \\ $$$$\left(\mathrm{1}\right)\:\mathrm{sin}^{\mathrm{2}} \:{x}\:−\:\mathrm{cos}^{\mathrm{2}} \:{x}\:=\:−\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{3}\:+\:\mathrm{2}{a}^{\mathrm{2}} \:−\:{a}^{\mathrm{4}} } \\ $$$$\left(\mathrm{2}\right)\:\mathrm{sin}^{\mathrm{4}} \:{x}\:+\:\mathrm{cos}^{\mathrm{4}} \:{x}\:=\:\frac{\mathrm{1}}{\mathrm{8}}\left(−\mathrm{3}{a}^{\mathrm{4}} \:+\:\mathrm{6}{a}^{\mathrm{2}} \:+\:\mathrm{5}\right) \\ $$$$\left(\mathrm{3}\right)\:\mathrm{sin}\:\mathrm{2}{x}\:=\:\frac{\mathrm{1}\:−\:{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\left(\mathrm{4}\right)\:\mathrm{tan}^{\mathrm{2}} \:{x}\:+\:\mathrm{cot}^{\mathrm{2}} \:{x}\:=\:\frac{−\mathrm{3}{a}^{\mathrm{4}} \:−\:\mathrm{2}{a}^{\mathrm{2}} \:+\:\mathrm{8}{a}\:+\:\mathrm{13}\:}{\mathrm{2}\left({a}^{\mathrm{4}} \:+\:\mathrm{2}{a}^{\mathrm{2}} \:+\:\mathrm{1}\right)} \\ $$

Commented by ajfour last updated on 26/Mar/17

$$\mathrm{none}\:\mathrm{seem}\:\mathrm{to}\:\mathrm{be}\:\mathrm{correct}.. \\ $$

Commented by Joel576 last updated on 26/Mar/17

$${why}? \\ $$

Commented by Joel576 last updated on 26/Mar/17

sorry, in statement (4) it must be tan^2 x + cot^2 x

$$\mathrm{sorry},\:\mathrm{in}\:\mathrm{statement}\:\left(\mathrm{4}\right)\:\mathrm{it}\:\mathrm{must}\:\mathrm{be}\:\mathrm{tan}^{\mathrm{2}} \:{x}\:+\:\mathrm{cot}^{\mathrm{2}} \:{x} \\ $$

Answered by ridwan balatif last updated on 26/Mar/17