Cos/Sin Problems

**qbkr21** · November 20th 2006, 03:38 PM

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)=?

**Soroban** · November 20th 2006, 04:55 PM

Hello, qbkr21!

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

**qbkr21** · November 20th 2006, 06:11 PM

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

**qbkr21** · November 20th 2006, 06:58 PM

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.

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