Simplify $\displaystyle (4 \cos^2 9^\circ -3)(4\cos^2 27^\circ -3 ) $ in terms of tangent.
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Originally Posted by shiny718 Simplify $\displaystyle (4 \cos^2 9^\circ -3)(4\cos^2 27^\circ -3 ) $ in terms of tangent. $\displaystyle \displaystyle \begin{align*} 1 + \tan^2{x} \equiv \frac{1}{\cos^2{x}} \end{align*}$
Sorry but I tried to use your hint but I couldn't get anywhere. Help please?
Originally Posted by shiny718 Sorry but I tried to use your hint but I couldn't get anywhere. Help please? Surely you can see from my hint that $\displaystyle \displaystyle \begin{align*} \cos^2{x} \equiv \frac{1}{1 + \tan^2{x}} \end{align*}$. Replace your square cosine terms with this.
Umm the correct answer I have in hand is $\displaystyle tan 9^\circ $. But I can't simplify all my way to this answer. Help me pls?
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