Cos/Sin Problems

• Nov 20th 2006, 03:38 PM
qbkr21
Cos/Sin Problems
None of the Problems can contain a Trig Identity...

cos(t)=-5/6

pi<t< 3pi/t

Give exact answers, do not use decimal numbers. The answer should be a fraction or an arithmetic expression. If the answer involves a square root it should be enter as sqrt; e.g. the square root of 2 should be written as sqrt(2).

Unknowns:

Sin(2t)=?
Cos(t/2)=?
sin(t/2)=?
• Nov 20th 2006, 04:55 PM
Soroban
Hello, qbkr21!

Quote:

Given: . $\cos(t) = -\frac{5}{6},\;\;\pi < t < \frac{3\pi}{2}$

Find: . $(a)\;\sin(2t)\qquad(b)\;\cos\left(\frac{t}{2}\righ t)\qquad(c)\;\sin\left(\frac{t}{2}\right)$

We are told that: . $\cos(t) \:=\:-\frac{5}{6}$

Since $\cos(t) \,=\,\frac{adj}{hyp}$, we know that: . $adj = -5,\;hyp = 6$

Pythagorus says: . $(opp)^2 + (adj)^2 \:=\:(hyp)^2$
. . so we have: . $(opp)^2 + (-5)^2 \:=\:6^2\quad\Rightarrow\quad(opp)^2 \:=\:11\quad\Rightarrow\quad opp = \pm\sqrt{11}$

Since $t$ is in quadrant 3, $opp = -\sqrt{11}$

So we have: . $\sin(t) \:=\:-\frac{\sqrt{11}}{6}\qquad \cos(t) \:=\:-\frac{5}{6}$

Now with a few identities, we can tackle the questions . . .

$(a)\;\sin(2t) \:=\:2\sin(t)\cos(t) \:=\:2\left(-\frac{\sqrt{11}}{6}\right)\left(-\frac{5}{6}\right) \:=\:\frac{5\sqrt{11}}{18}$

$(b)\;\cos\left(\frac{t}{2}\right) \:=\:\pm\sqrt{\frac{1 + \cos t}{2}} \:= \:\pm\sqrt{\frac{1 - \left(-\frac{5}{6}\right)}{2}} \:=\:\pm\sqrt{\frac{1}{12}} \:=\:\pm\frac{\sqrt{3}}{6}$

. . .Since $t$ is in quadrant 3, $\frac{t}{2}$ is in quadrant 2 where cosine is negative.

. . .Therefore: . $\cos\left(\frac{t}{2}\right)\;=\;-\frac{\sqrt{3}}{6}$

$(c)\;\sin\left(\frac{t}{2}\right) \:=\:\pm\sqrt{\frac{1-\cos(t)}{2}} \:=\:\pm\sqrt{\frac{1-\left(-\frac{5}{6}\right)}{2}} \:=\:\pm\sqrt{\frac{11}{12}}\:=\:\pm\frac{\sqrt{33 }}{6}$

. . .Since $t$ is in quadrant 3, $\frac{t}{2}$ is in quadrant 2 where sine is positive.

. . .Therefore: . $\sin\left(\frac{t}{2}\right) \;=\;\frac{\sqrt{33}}{6}$

• Nov 20th 2006, 06:11 PM
qbkr21
RE:Thanks so Much!!
Thanks Soroban you guys are such brains...really am baffeled by how much you guys know, no one needs to know this much math, this is all crazy!!!

Thanks
• Nov 20th 2006, 06:58 PM
qbkr21
Re:
Dude I don't think you realize how much easier you have made this for me. I have but in 20 plus hours over the past 2 days, and for the first time since reading you post this is now clicking. Gosh I really, really thank you for help.