# Some trig problems =/

• Dec 19th 2006, 05:51 PM
aargh27
Some trig problems =/
I have some work and im confused on these problems... I did all i can (looked in book and such) and need help.

Simplify the expression cos (X-5pie /2)
For this one i know that you can go to (cosx)(cos-5pie/2) + (sinx)(sin-5pie/2) But i dont know what to do from there

second is If cotx= 4/3 and 180degrees< x < 270degrees, find the exact values of sin2x and cos2x
Here, i know if there was a triangle, there would be the 5,4,and3 to it, but i dont know how you go to the sin2x and cos2x

third is Find the solutions of sin2x = cosx that are in interval [0, 2pie)

If anyone could help, youd be my hero. I have 8 tests this week and im stressing out on these problems....
• Dec 19th 2006, 06:52 PM
Soroban
Hello, aargh27!

Quote:

1) Simplify: .$\displaystyle \cos\left(x - \frac{5\pi}{2}\right)$

What you did was almost correct . . .

$\displaystyle \cos\left(x - \frac{5\pi}{2}\right)\;=\;\cos x\cdot\cos\frac{5\pi}{2} + \sin x\cdot\sin\frac{5\pi}{2}$

Note that $\displaystyle \frac{5\pi}{2}$ is equivalent to $\displaystyle \frac{\pi}{2}$
. . so we have: .$\displaystyle \cos x\cos\frac{\pi}{2} - \sin x\sin\frac{\pi}{2}$

. . . . . . . . . .$\displaystyle = \;\cos x\cdot0 - \sin x\cdot 1 \:=\:\boxed{-\sin x}$

Quote:

2) If $\displaystyle \cot x= \frac{4}{3}$ and $\displaystyle 180^o < x < 270^o$degrees,
find the exact values of $\displaystyle \sin2x$ and $\displaystyle \cos2x$

First, note that $\displaystyle x$ is in Quadrant 3.

Since $\displaystyle \cot x = \frac{adj}{opp}$, we have: .$\displaystyle opp =-3,\;adj = -4\quad\Rightarrow\quad hyp = 5$
. . Then: .$\displaystyle \sin x = -\frac{3}{5},\;\cos x = -\frac{4}{5}$

Using double-angle identities, we have:

. . $\displaystyle \sin2x\;=\;2\!\cdot\!\sin x\!\cdot\!\cos x \;=\;2\left(-\frac{3}{5}\right)\left(-\frac{4}{5}\right) \;=\;\boxed{\frac{24}{25}}$

. . $\displaystyle \cos2x\;=\;\cos^2x - \sin^2x\;=\;\left(-\frac{4}{5}\right)^2 - \left(-\frac{3}{5}\right)^2\;=\;\boxed{\frac{7}{25}}$

Quote:

3) Find the solutions of: .$\displaystyle \sin2x \:= \:\cos x$ in the interval $\displaystyle [0,2\pi)$

Replace $\displaystyle \sin2x$ with $\displaystyle 2\sin x\cos x$

. . $\displaystyle 2\sin x\cos x \:=\:\cos x\quad\Rightarrow\quad 2\sin x\cos x - \cos x \:=\:0$

Factor: .$\displaystyle \cos x(2\sin x - 1)\:=\:0$

And we have two equations to solve:

. . $\displaystyle \cos x\:=\:0\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}$
. . $\displaystyle 2\sin x - 1\:=\:0\quad\Rightarrow\quad\sin x = \frac{1}{2}\quad\Rightarrow\quad\boxed{x \:=\:\frac{\pi}{6},\:\frac{5\pi}{6}}$

• Dec 20th 2006, 03:31 PM
aargh27
Ty again for your help. However just one thing that I found as these were graded. For the first problem it was +sinx .. not -sinx :p