Question

find exact value of sin(x+pi/6) if sinx=-12/13 and tan<0

Answer 1

Expand sin(x+pi/6) using sum identity for sin
You will also need to find cos(x)
Recall \[\cos^2(x)=1-\sin^2(x) \\cos(x)=\pm \sqrt{1-\sin^2(x)}\]
You will need to determine if cos(x)>0 of if cos(x)<0.
You can do find knowing that sin is negative and tangent is negative, forcing cosine to be?

Answer 2

positive

Answer 3

So would I use the equation sin(pi/3+pi/2)=sin(pi/3)cos(pi/2)+cos(pi/3)sin(pi/2)?

Answer 4

or would it be sin(x+pi/6)=sin(-12/13)cos(11pi/6)+cos(-12/13)sin(11pi/6)?

Answer 5

sin(x+pi/6)
=sin(x)cos(pi/6)+sin(pi/6)cos(x)

Answer 6

you are not given x is -12/13
you are given sin(x) is -12/13

Answer 7

I gave you an equation to find cos(x)

Answer 8

is this correct? sin(-12/13)cos(sqrt3/2)+cos(5/13)sin(1/2)

Answer 9

Don't replace x with -12/13
replace sin(x) with -12/13
and I don't think x is 5/13

Answer 10

cos(pi/6) is sqrt(3)/2 not cos(sqrt(3)/2)

Answer 11

sin(pi/6)=1/2 not sin(1/2)

Answer 12

\[\sin(x)=\frac{-12}{13} \\ \cos(x)=\sqrt{1-(\frac{-12}{13})^2}=\sqrt{1-\frac{144}{169}}=\sqrt{\frac{169-144}{169}}=\sqrt{\frac{25}{169}}=\frac{5}{13}\]

Answer 13

So this tells us cos(x) is 5/13
not that x is 5/13

Answer 14

\[\sin(x+\frac{\pi}{6})=\sin(x)\cos(\frac{\pi}{6})+\cos(x)\sin(\frac{\pi}{6}) \\ =\frac{-12}{13} \cdot \frac{\sqrt{3}}{2}+\frac{5}{13} \cdot \frac{1}{2}\]

Answer 15

Finish simplifying that.

Answer 16

That is what I did and I came up with -12sqrt3+5/26

Answer 17

Ok well you were replacing x with -12/13 and 5/13
even though it is already weird to say x is taking on two different values at the same time
So that was very concerning...

Answer 18

And the other weird thing is cos(pi/6) =sqrt(3)/2 not cos(sqrt(3)/2))
and sin(pi/6)=1/2 not sin(1/2)

Answer 19

I think I was just typing it differently which made it look confusing?

Answer 20

Not just differently but incorrectly.

Answer 21

Don't let your teacher see that way. :p

Answer 22

ok thank you