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Prove the identity $\sin^4α-\cos^4α=2\sin^2α-1$

Well, I thought to start it this way: $$(\sin^2α-\cos^2α)(\sin^2α+\cos^2α)=2\sin^2α-1=>\\(\sin α-\cos α)(\sin α+\cos α)(\sin^2α+\cos^2α)=2\sin^2α-1$$ I don't know how to continue...

Thomas Andrews
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seda
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    Hint $cos^2a-sin^2a=cos2a$ and $sin^2a+cos^2a=1$ Gets you almost there – imranfat Mar 10 '15 at 17:18
  • You were almost finished. You got to $(\sin^2\alpha+\cos^2\alpha)(\sin^2\alpha-\cos^2\alpha)$. This is $\sin^2\alpha-\cos^2\alpha$. Now replace $\cos^2\alpha$ by $1-\sin^2\alpha$. – André Nicolas Mar 10 '15 at 17:22

2 Answers2

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Hint: $ \sin^2 \alpha + \cos^2\alpha=1 $ substituting in the two sides you have the identity.

N. F. Taussig
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Emilio Novati
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$\cos^4a-\sin^4a=(\cos^2a−\sin^2a)(\cos^2a+\sin^2a)=(2\cos^2a-1)*1$
(becouse $\cos^2a−\sin^2a=\cos^2a-(1-\cos^2a)=2\cos^2a-1$)

Antony
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