\begin{align*} cos36=1-2sin^2(18) \end{align*}

$cos 36=sin 54$

oh come on really!

cos 36 - cos 54 = - 2sen 18.sen54 = sen 36.sen108\ 2.cos18.cos54 = 1\2

marcilio wrote: cos 36 - cos 54 = - 2sen 18.sen54 = sen 36.sen108\ 2.cos18.cos54 = 1\2 Math_passion wrote: \begin{align*} cos36=1-2sin^2(18) \end{align*} vjdjmathaddict wrote: oh come on really! Please a solution in details, im a beginner on trigonometry

Please move this high school math forum.

Duarti wrote: marcilio wrote: cos 36 - cos 54 = - 2sen 18.sen54 = sen 36.sen108\ 2.cos18.cos54 = 1\2 Math_passion wrote: \begin{align*} cos36=1-2sin^2(18) \end{align*} vjdjmathaddict wrote: oh come on really! Please a solution in details, im a beginner on trigonometry Solve it by quadratic equation

Solve it like this: $cos 36-sin 18=\frac {(2 cos 36 cos 18-2sin 18 cos 18)}{2 cos 18}=\frac {(cos 18+cos 54-sin 36)}{2 cos 18}=\frac {1}{2} $

Math_passion wrote: Duarti wrote: marcilio wrote: cos 36 - cos 54 = - 2sen 18.sen54 = sen 36.sen108\ 2.cos18.cos54 = 1\2 Math_passion wrote: \begin{align*} cos36=1-2sin^2(18) \end{align*} vjdjmathaddict wrote: oh come on really! Please a solution in details, im a beginner on trigonometry Solve it by quadratic equation Why to use quadratic in this one liner!!

Because in the will be only one root

Saptarshi17_math wrote: Solve it like this: $cos 36-sin 18=\frac {(2 cos 36 cos 18-2sin 18 cos 18)}{2 cos 18}=\frac {(cos 18+cos 54-sin 36)}{2 cos 18}=\frac {1}{2} $ See this!

but i computed value of sin18

Math_passion wrote: but i computed value of sin18 No need. In exam you may not always be able to find out like you computed.

Sorry, but i didn't ask you? In this problem this is computeble.

Found a nice one, but please correct if I messed up somewhere. Draw an obtuse triangle with angles A(126 degrees) B(36 degrees) and C(18 degrees) and sides AB (2) and AC(1). Drop the altitude from angle A unto side BC, and you get two right triangles. Express the altitude in terms of sin/cos of both sides then proceed.

Duarti wrote: Prove that : $\cos36-\sin18=\frac{1}{2}$

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