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If sin(...)
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\[\LARGE \sin(\sin^{-1}\frac{1}{5}+\cos^{-1}x)=1,~find~the~value~of~x\]
1/5
thisi s a trick
HOW :O
#genius!
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lol\[ \sin^{-1}{1\over5}+\left({\pi\over2}-\sin^{-1}(x)\right)={\pi\over2} \]
:/ lol
I see what you did there,but how did you simplify it?
\[ \sin^{-1}{1\over5}+\left(\cancel{\pi\over2}-\sin^{-1}(x)\right)=\cancel{\pi\over2} \]
oh great :D
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so \[\large \sin^{-1}x=\frac{1}{5}\] same for cos?
what! \[\sin^{-1}x=\sin^{-1}{1\over5}\implies x={1\over 5}\]
:/ silly me :P
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