if sinx = 4/5 and x is in the first quadrant, find sin2x What I know : sin2x = 2sinxcosx (in my notes) cos2x = 1-2sin^2x = -7/25 (I solved that myself) sinx = 4/5 (given)
hint: we have: \[\Large \cos x = \sqrt {1 - {{\left( {\sin x} \right)}^2}} = \sqrt {1 - \frac{{16}}{{25}}} = ...?\]
that confused me even more lol
Michael i explained to her the whole thing step by step and she didnt get it 😂
I try to explain another time @joyraheb from my formula above, I get this: \[\Large \begin{gathered} \cos x = \sqrt {1 - {{\left( {\sin x} \right)}^2}} = \sqrt {1 - \frac{{16}}{{25}}} = \hfill \\ \hfill \\ = \sqrt {\frac{{25 - 16}}{{25}}} = \frac{3}{5} \hfill \\ \end{gathered} \] so, substituting, I can write: \[\Large \sin \left( {2x} \right) = 2\sin x\cos x = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = ...?\] Please complete @jxaf
the easy solution to find cos(x) is to draw the 1st quadrant triangle |dw:1440279636760:dw|
***if sinx = 4/5 and x is in the first quadrant, find sin2x **** ****sin2x = 2sinxcosx (in my notes)*** from your notes you have a way to find sin2x. you need to know sinx (which you do know). you also need cos x. so that is the problem. how to find cos x, so then you can use your formula. I always try to draw a picture, and then use pythagoras (I hope you remember c^2 = a^2 + b^2 ) ? first quadrant. sin is opp/hyp it looks just like how campbell drew it |dw:1440281648678:dw|
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