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(6cos\({^2}\)24°-6sin\({^2}\)24°)tan48°

 

I used the double angle identity cos2A = cos\({^2}\)A-sin\({^2}\)

 

and I got 6cos2(24)°(tan48°)

 

I could change the tan to sin/cos but then I'm not sure what to do next?

 

A hint would be great! :)

 Feb 21, 2019
 #1
avatar+21848 
+5

(6cos224°-6sin224°)tan48°

 

\(\begin{array}{|rcll|} \hline &&\mathbf{ (6\cos^224^\circ-6\sin^224^\circ)\tan48^\circ } \\\\ &=& 6\cdot (\cos^224^\circ-\sin^224^\circ)\tan(48^\circ) \quad | \quad \cos(48^\circ) = \cos^2(24^\circ) - \sin^2(24^\circ) \\\\ &=& 6\cdot \cos(48^\circ)\cdot \tan(48^\circ) \quad | \quad \tan(48^\circ) = \dfrac{\sin(48^\circ)}{\cos(48^\circ)} \\\\ &=& 6\cdot \cos(48^\circ)\cdot \dfrac{\sin(48^\circ)}{\cos(48^\circ)} \\\\ &=& 6\cdot \dfrac{\sin(48^\circ)\cos(48^\circ)}{\cos(48^\circ)} \\\\ &\mathbf{=}& \mathbf{6\cdot \sin(48^\circ)} \\ \hline \end{array} \)

 

laugh

 Feb 21, 2019
edited by heureka  Feb 21, 2019
 #2
avatar
+1

I'm just wondering how was my approach incorrect?

 

** Isn't sine and cosine squared?

Guest Feb 21, 2019
edited by Guest  Feb 21, 2019
 #3
avatar+21848 
+4

yes, sorry

but the calculation is okay.

 

laugh

heureka  Feb 21, 2019
 #4
avatar+99301 
+2

\((6cos^2 24°-6sin^2 24°)tan48°\)

 

I used the double angle identity \(cos2A = cos^2 A-sin^2 A \)

 

and I got \(6cos2(24)°(tan48°)\)

 

I could change the tan to sin/cos but then I'm not sure what to do next?

 

--------

\(6cos[\color{red}{2(24)}\color{black}{]°(tan48°)}\\ =6cos[\color{red}{48}\color{black}{]°(tan48°)}\\ =6cos[\color{red}{48}\color{black}{]°\cdot\frac{sin48}{cos48}}\\ =6sin48\)

.
 Feb 21, 2019

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