Not Your Typical Trigonometric Values

$\large \sin 20^\circ-\cos 20^\circ=L$ (Squaring and using $\color{#D61F06}{\sin^2 \theta+\cos^2 \theta=1}$ ) $\large \implies 2\sin 20^\circ\cos 20^\circ=\sin 40^\circ=1-L^2$ $~~~~~~~~~\left(\color{#D61F06}{2\sin \theta \cos\theta=\sin 2\theta}\right)$ $\large \therefore\cos 40^\circ=\sqrt{1-(\sin 40^\circ)^2}$ $\large =\sqrt{1-(1-L^2)^2}$ $~~~~~~~~\color{#D61F06}{Use ~a^2-b^2=(a+b)(a-b)}$ $\large=\sqrt{(L^2)(2-L^2)}$ $\Large =\color{#0C6AC7}{-L\sqrt{2-L^2}}$

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$\color{#20A900}{Note:-}\text{Since sin 20°<cos 20°, hence L<0}$ $\cos 40^\circ >0 \implies \cos 40^\circ= -L\sqrt{2-L^2}$ $~~~~~~~~~~~~~~\text { (And not } L\sqrt{2-L^2})$

Sin(20)-cos(20)=L

(Sin(20)-cos(20))^2+(sin(20)+cos(20))^2=2 Therefore

Sin(20)+cos(20)=√(2-L^2)

Cos(40)=cos^2(20)-sin^2(20)

=-(sin(20)-cos(20))×(cos(20)+sin(20))

= -L×√(2-L^2)