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Tomtit [17]
3 years ago
8

Is 2 feet longer than 30 inches

Mathematics
1 answer:
loris [4]3 years ago
5 0
1 foot is 12 inches. 2 feet is 24 inches. yea so I'd say 30 inches is definitely longer
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\sin(a-b)=\sin a \cos b-\cos a \sin b. If we let a=\cos^{-1} \left(\frac{5}{6} \right) and b=\tan^{-1} \left(\frac{1}{2} \right), then the given expression is equal to:

\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)

Using the Pythagorean identities \sin^{2} x+\cos^{2} x=1 and \tan^{2} x+1=\sec^{2} x,

1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2}  \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}

2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}

\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}

This means we can write the original expression as:

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