Question

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)

Answer 1

hint:
we have:
\[\Large \cos x = \sqrt {1 - {{\left( {\sin x} \right)}^2}} = \sqrt {1 - \frac{{16}}{{25}}} = ...?\]

Answer 2

that confused me even more lol

Answer 3

Michael i explained to her the whole thing step by step and she didnt get it 😂

Answer 4

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

Answer 5

the easy solution to find cos(x) is to draw the 1st quadrant triangle

Answer 6

***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