Question

مطلوب است محاسبه: $$cos(x-y)=?$$ در صورتی که: $$ \frac{log(sinx)+log(cosx)}{log(siny)-log(cosy)} =3 \wedge x+y= \frac{ \pi }{2} \wedge (x,y > 0)$$

Answer 1

$\frac{\text{log}(\text{sin}x)+\text{log}(\text{cos}x)}{\text{log(sin}y)-\text{log(cos}y)} =3\implies \frac{\text{log(sin}x\text{cos}x)}{\text{log(cot}x)}=3\implies \text{sin}x\text{cos}x=\frac{\text{cos}^{3}x}{\text{sin}^{3}x}=\text{sin}x\text{cos}x\implies \text{cos}^{2}x=\text{sin}^{4}x\implies \text{sin}^{4}x+\text{sin}^{2}x-1=0\implies \text{sin}^{2}x=\frac{\sqrt{5}-1}{2}\implies \text{cos}^{2}x=\frac{3-\sqrt{5}}{2}$ $\text{cos}(x-y)=\text{sin}2x=\sqrt{4\text{sin}^{2}x\text{cos}^{2}x}=\sqrt{(\sqrt{5}-1)(3-\sqrt{5})}$

Answer 2

$$x+y= \frac{ \pi }{2} \Longrightarrow tany= \frac{1}{tanx} \wedge cos(x-y)=cos (2x- \frac{a}{b} )=sin2x \Longrightarrow log(sinx.cosx)=3log \frac{siny}{cosy} \Longrightarrow log(sinx.cosx)=log tan^{3} y \Longrightarrow sin x.cosx= \frac{1}{ tan^{3}x } \Longrightarrow \frac{sin2x}{2} = \frac{1}{ tan^{3}x } \Longrightarrow \frac{tanx}{1+ tan^{2} x} = \frac{1}{ tan^{3} x} \Longrightarrow tan^{4} x- tan^{2}x -1=0 \Longleftrightarrow tan^{2}x= \frac{1+ \sqrt{5} }{2} \Longrightarrow cos 2x= \frac{1- tan^{2} x}{1+ tan^{2} x} = \frac{1- \sqrt{5} }{3+ \sqrt{5} } \Longrightarrow cos 2x=2- \sqrt{5} \Longrightarrow sin 2x =2 \sqrt{ \sqrt{5} -2} $$