help with a pre-calculus problem?

Question

find the exact value of sin ( theta + pi/6) , given that
cos(theta) = -3/4 and 0 ≤theta≤ pi
if you could show work that would be awesome
I really appreciate your help guys

Answer 1

if cos Θ = -3/4, Θ is in quadrant 2 and sin Θ = √7/4. then

sin(Θ + π/6) =
sin Θ cos π/6 + cos Θ sin π/6 =
(√7/4)(√3/2) + (-3/4)(1/2) =
(√21 - 3)/8

Answer 2

Hi,

If cos(theta) = -3/4, then theta is in the second quadrant. -3 is the adjacent side, 4 is the hypotenuse. Using the Pythagorean theorem, the sqrt(7) is the length of the opposite side. So, sin(theta) = sqrt(7)/4

Since sin (A + B) = (sin A)(cos B) + (cos A)(sin B). then
sin (theta + Pi/6) = (sin theta)(cos Pi/6) + (cos theta)(sin Pi/6)

We know that cos( Pi/6) = sqrt(3)/2 and sin(Pi/6) = 1/2.

Plugging in values gives:

sin(theta + Pi/6) = (sin theta)(cos Pi/6) + (cos theta)(sin Pi/6)
sin(theta + Pi/6) = [sqrt(7)/4][sqrt(3)/2] + (-3/4)(1/2) =
sqrt(21)/8 - 3/8 = [sqrt(21) - 3]/8

I hope that helps!!! :-)

Answer 3

The third leg of the right triangle would be sqrt(16-9)=sqrt(7). Therefore, sin(theta)=sqrt(7)/4.
Cos(pi/6)= sqrt(3)/2 and Sin(pi/6)=1/2
Therefore, the answer is cos(theta)*cos(pi/6)-sin(theta)*sin(pi/6)= (-3/4)*sqrt(3)/2-sqrt(7)/4*1/2

Answer 4

Use your sum-of-sines law to express sin(theta+pi/6). You already know sin(pi/6) and cos(pi/6), so now all you have is terms involving sin(theta) & cos(theta). You also know sin(theta) in terms of cos(theta) from your basic trig identities. Make the substitution so you have only cos(theta) terms left, and then plug in -3/4 for those.

Other than that, you're on your own -- I'm not going to do the work for you. That's cheating.

Answer 5

To find the exact value of sin(t + &pi/6) given that cos(t) = -3/4, we must first use the sine addition identity which goes as follows.

sin(a + b) = sin(a)cos(b) + sin(b)cos(a)

That means
sin(t + &pi/6) = sin(t)cos(&pi/6) + sin(&pi/6)cos(t)
= sin(t) (sqrt(3)/2) + (1/2)cos(t)

But cos(t) = -3/4, so we have

= sin(t) (sqrt(3)/2) + (1/2)(-3/4)
= sin(t) (sqrt(3)/2) - 3/8

To get sin(t), we use the fact that cos(t) = -3/4.
Use right angles and SOHCAHTOA to get sin(t).

cos(t) = -3/4 = adj/hyp
adj = -3
hyp = 4, so by Pythagoras,
opp = sqrt(4^2 - (-3)^2) = sqrt(16 - 9) = sqrt(6)

sin(t) = opp/hyp = sqrt(6)/4

= sin(t) (sqrt(3)/2) - 3/8
= (sqrt(6)/4)(sqrt(3)/2) - 3/8
= sqrt(18)/8 - 3/8
= 3sqrt(2)/8 - 3/8
= [3sqrt(2) - 3]/8
= (3/8) (sqrt(2) - 1)