Track / Stage 5 / 345 of 400 #945 of 1964
Problem 945 AIME late Difficulty 5.8 Prove it
3.039. cos 4 α − sin 4 α ctg 2 α = − 1 \cos 4 \alpha-\sin 4 \alpha \operatorname{ctg} 2 \alpha=-1 cos 4 α − sin 4 α ctg 2 α = − 1
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3.039. cos 4 α − sin 4 α cot 2 α = − 1 \cos 4 \alpha-\sin 4 \alpha \cot 2 \alpha=-1 cos 4 α − sin 4 α cot 2 α = − 1
This one wants a proof. Work it on paper, then read the official solution and mark
yourself. Be honest about it: the record is only any use to you if it is.
I solved it I didn't Skip
Official solution Solution.
cos 4 α − sin 4 α ctg 2 α = cos 4 α − sin 4 α ⋅ cos 2 α sin 2 α = = sin 2 α cos 4 α − cos 2 α sin 4 α sin 2 α = sin ( − 2 α ) sin 2 α = − sin 2 α sin 2 α = − 1
\begin{aligned}
& \cos 4 \alpha-\sin 4 \alpha \operatorname{ctg} 2 \alpha=\cos 4 \alpha-\sin 4 \alpha \cdot \frac{\cos 2 \alpha}{\sin 2 \alpha}= \\
& =\frac{\sin 2 \alpha \cos 4 \alpha-\cos 2 \alpha \sin 4 \alpha}{\sin 2 \alpha}=\frac{\sin (-2 \alpha)}{\sin 2 \alpha}=\frac{-\sin 2 \alpha}{\sin 2 \alpha}=-1
\end{aligned}
cos 4 α − sin 4 α ctg 2 α = cos 4 α − sin 4 α ⋅ sin 2 α cos 2 α = = sin 2 α sin 2 α cos 4 α − cos 2 α sin 4 α = sin 2 α sin ( − 2 α ) = sin 2 α − sin 2 α = − 1
The identity is proven.
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