## anonymous one year ago 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)

1. Michele_Laino

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

2. anonymous

that confused me even more lol

3. anonymous

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

4. Michele_Laino

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

5. campbell_st

the easy solution to find cos(x) is to draw the 1st quadrant triangle |dw:1440279636760:dw|

6. phi

***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|

7. phi

we know (I hope!) cos x is adj/hyp so if we can find the adjacent side, we can do this problem... a^2 + 4^2 = 5^2 or a^2+16=25 a^2= 25-16 a^2= 9 a= 3 (btw, try to memorize 3,4,5 is a right triangle. It shows up enough times in problems) so cos x = 3/5 now you can do sin(2x)= 2* sinx * cos x = $$2\cdot \frac{4}{5}\cdot \frac{3}{5} = \frac{24}{25}$$

8. anonymous

Ooooh , using the triangle makes A LOT of sense . Thank you ! and sorry , I had to do something with my partner for school