Recent content by Silver Bolt

Corrected now

$\sin\left({A}\right)+\sin\left({A+\frac{2\pi}{3}}\right)+\sin\left({A+\frac{4\pi}{3}}\right)=0$...

$\sin^8\left({A}\right)-\cos^8\left({A}\right)=(\sin^2\left({A}\right)-\cos^2\left({A}\right)(1-2\sin^2\left({A}\right)\cos^2\left({A}\right))$ $L.H.S=(\sin^2\left({A}\right)-\cos^2\left({A}\right)(1-2\sin^2\left({A}\right)\cos^2\left({A}\right))$ $...

$\frac{\cot\left({A}\right)\cos\left({A}\right)}{\cot\left({A}\right)+\cos\left({A}\right)}=\frac{\cot\left({A}\right)-\cos\left({A}\right)}{\cot\left({A}\right)\cos\left({A}\right)}$ $L.H.S=\frac{\cot\left({A}\right)\cos\left({A}\right)}{\cot\left({A}\right)+\cos\left({A}\right)}$...

$\frac{1-\tan\left({A}\right)}{1+\tan\left({A}\right)}=\frac{\cot\left({A}\right)-1}{\cot\left({A}\right)+1}$ $L..H.S=\frac{1-\frac{\sin\left({A}\right)}{\cos\left({A}\right)}}{1+\frac{\sin\left({A}\right)}{\cos\left({A}\right)}}$...

$\sin^4(A)-\sin^2(A)\cos^2(A)+\cos^4(A)=(\sin^2(A)+\cos^2(A))^2-3\sin^2(A)\cos^2(A)$ This part can be rewritten as $\sin^4(A)+\cos^4(A)-\sin^2(A)\cos^2(A)=(\sin^2(A)+\cos^2(A))^2-\sin^2(A)\cos^2(A)$ Expanding $(\sin^2(A)+\cos^2(A))^2=\sin^4(A)+2\sin^2(A)\cos^2(A)+cos^4(A)$ Now using it...

A little explanation on how did that -3 come in $\sin^2(A)+\cos^2(A))^2-3\sin^2(A)\cos^2(A)$

Prove $Cos^6A+Sin^6A=1-3 \hspace{0.2cm}Sin^2 A\hspace{0.02cm}Cos^2A$ So far, $Cos^6A+Sin^6A=1-3 \hspace{0.2cm}Sin^2 A\hspace{0.02cm}Cos^2A$ $L.H.S=(Cos^2A)^3+(Sin^2A)^3$ $=(Cos^2A+Sin^2A)(Cos^4A-Cos^2ASin^2A+Sin^4A)$...