Question

If sinθ=-1/4 and θ terminates in third quad., find the exact value of sin 2θ

Answer 1

???

Answer 2

Okie, found it :)

Answer 3

\[\sin²(\theta)=1/16=[1-\cos(2.\theta)]/2\]
\[\cos(2.\theta)=1-1/8\]
\[\sin(2.\theta)=-\sqrt{1-\cos(2.\theta)^{2}}\]

Answer 4

sin2x = 2sinxcosx
cos^2x = 1 - sin^2x = 1 - ( -1/4)^2

Answer 5

Can you follow?

Answer 6

the "-" is due to third quad

Answer 7

@victor do you need more help?

Answer 8

i dont understand how you did that?

Answer 9

i understand sin2x = 2sinxcosx but then how did you get cos^2x = 1 - sin^2x = 1 - ( -1/4)^2 @Chlorophyll

Answer 10

Given: sinx, and theta in 3rd quadrant
Ask : sin2x = 2sinxcosx

Answer 11

=> need cosx

Answer 12

Sin^2x + cos^2x = 1

Answer 13

ohhh i see it. so youre not using the double angle you just replaced it

Answer 14

cosx = sqrt ( 1 - sin^2x )
I don't want to confuse you, so I go step by step!

Answer 15

Are you okie up to this step?

Answer 16

yea cosx=1+sin(1/2)

Answer 17

NO, that's different formula, half angle!

Answer 18

cosx = sqrt ( 1 - sin^2x )
You want angle is x not x/2, not 2x

Answer 19

yea but isnt x -1/4 so sqrt 1/4 is 1/2

Answer 20

cosx = sqrt ( 1 - sin^2x )
= sqrt ( 1 - 1/16 )
= sqrt ( 15/16) = sqrt (15) / 4

Answer 21

oh i sqrt first. you cant just leave it as sqrt 15/16

Answer 22

and then you plug in your sqrt to cos

Answer 23

That's how you find cosx from sinx ( very traditional way)

Answer 24

so whats the final answer?

Answer 25

Notice that x belongs the THIRD quadrant -> both cosx and sinx negative
=> cosx = - sqrt (15)/4

Now plug it into sin2x = 2 sinx * cosx
= ....

Answer 26

I assume you're fine here, all you do is multiply 2 sinx * cosx

Answer 27

\[\sqrt{15/4} or \sqrt{15/16}\]

Answer 28

for cos

Answer 29

\[cosx = \sqrt{15}/ 4\]

Answer 30

\[\sin (2 \theta )=2 \sin (\theta ) \cos
(\theta )=2 \sin (\theta ) \sqrt{1-\sin
^2(\theta )}=\frac{2}{4} (-1)
\sqrt{1-\frac{1}{16}}=-\frac{\sqrt{15}}{
8}
\]