# Algebraically solving sin and cos functions, helpl please!

• Oct 18th 2012, 06:17 PM
skg94
Algebraically solving sin and cos functions, helpl please!
i cant find any symbols on this site so if its confusing please do tell.

sin[2(x-(pi/4))]=sq root2/2 for 0< x <2pi and for all feta (x=feta)

my attempt:
2(x-(pi/4))= pi/4 , 7pi/4
x-(pi/4) =pi/8 , 7pi/8
x = 3pi/8 9pi/8

based on the domain , only 3pi/8 fits.

for all feta = 3pi/8 +pi(n) , 9pi/8 + pi(n) where n is an element of the integers.

4cos(x+30degrees)+5=7 for 0<x<360 and for all feta (x=feta)

COS(X+30)=1/2
x+30 = 1.047197551

i dont know where to go from here

2sin[2(x-(pi/6))]=-sqroot3 for 0<x<2pi and for all feta (x=feta)
sin[2(x-(pi/6))] = -sqroot3/2
2(x-(pi/6))= pi/3 5pi/3
x-(pi/6) = pi/6 ,5pi/6
x= pi/3 and pi (both work)

for all: pi +pi(N) , pi/3 + pi(n) where n is an element of the integers.

3sin(x-1)=2 0<x<360 round to the tenth of a degree and for all feta (x=feta)
sin(x-1)=2/3
x-1=.7297276567

again im having problems with these decimal and degrees ones

and if you can check over the two i finished please and thank you , if i made mistakes please point them out even the obvious ones

• Oct 18th 2012, 08:05 PM
Prove It
Re: Algebraically solving sin and cos functions, helpl please!
Quote:

Originally Posted by skg94
i cant find any symbols on this site so if its confusing please do tell.

sin[2(x-(pi/4))]=sq root2/2 for 0< x <2pi and for all feta (x=feta)

my attempt:
2(x-(pi/4))= pi/4 , 7pi/4
x-(pi/4) =pi/8 , 7pi/8
x = 3pi/8 9pi/8

based on the domain , only 3pi/8 fits.

for all feta = 3pi/8 +pi(n) , 9pi/8 + pi(n) where n is an element of the integers.

4cos(x+30degrees)+5=7 for 0<x<360 and for all feta (x=feta)

COS(X+30)=1/2
x+30 = 1.047197551

i dont know where to go from here

2sin[2(x-(pi/6))]=-sqroot3 for 0<x<2pi and for all feta (x=feta)
sin[2(x-(pi/6))] = -sqroot3/2
2(x-(pi/6))= pi/3 5pi/3
x-(pi/6) = pi/6 ,5pi/6
x= pi/3 and pi (both work)

for all: pi +pi(N) , pi/3 + pi(n) where n is an element of the integers.

3sin(x-1)=2 0<x<360 round to the tenth of a degree and for all feta (x=feta)
sin(x-1)=2/3
x-1=.7297276567

again im having problems with these decimal and degrees ones

and if you can check over the two i finished please and thank you , if i made mistakes please point them out even the obvious ones

\displaystyle \begin{align*} \sin{\left[ 2\left( x - \frac{\pi}{4} \right) \right]} &= \frac{\sqrt{2}}{2} \\ 2\left( x - \frac{\pi}{4} \right) &= \left\{ \frac{\pi}{4}, \pi - \frac{\pi}{4} \right\} + 2\pi n \textrm{ where } n \in \mathbf{Z} \\ 2\left( x - \frac{\pi}{4} \right) &= \left\{ \frac{\pi}{4}, \frac{3\pi}{4} \right\} + 2\pi n \\ x - \frac{\pi}{4} &= \frac{1}{2} \left( \left\{ \frac{\pi}{4}, \frac{3\pi}{4} \right\} + 2\pi n \right) \\ x - \frac{\pi}{4} &= \left\{ \frac{\pi}{8}, \frac{3\pi}{8} \right\} + \pi n \\ x &= \left\{ \frac{\pi}{8} + \frac{\pi}{4}, \frac{3\pi}{8} + \frac{\pi}{4} \right\} + \pi n \\ x &= \left\{ \frac{3\pi}{8}, \frac{5\pi}{8} \right\} + \pi n \\ x &= \frac{3\pi}{8}, \frac{5\pi}{8}, \frac{11\pi}{8}, \frac{13\pi}{8} \textrm{ in the interval } 0 \leq x \leq 2\pi \end{align*}
• Oct 18th 2012, 08:20 PM
skg94
Re: Algebraically solving sin and cos functions, helpl please!
well thank you for that, could you potentially help me with the others? please that helped me alot

how do i know if my second angle should equal to pi or 2pi? based on the period? at the start

where you did 3pi/4 and not 7pi/4
• Oct 18th 2012, 08:32 PM
Prove It
Re: Algebraically solving sin and cos functions, helpl please!
Quote:

Originally Posted by skg94
well thank you for that, could you potentially help me with the others? please that helped me alot

how do i know if my second angle should equal to pi or 2pi? based on the period? at the start

where you did 3pi/4 and not 7pi/4

I got \displaystyle \begin{align*} \frac{3\pi}{4} \end{align*} from the fact that the sine function is positive in the SECOND quadrant, not the fourth.

You can tell how many solutions there would be based on the period. You know that there should be two solutions in each period, and since the period was \displaystyle \begin{align*} \frac{2\pi}{2} = \pi \end{align*}, there are two periods in the region \displaystyle \begin{align*} 0 \leq x \leq 2\pi \end{align*}, and so there must be four solutions in that region. I find it easiest to add \displaystyle \begin{align*} 2\pi n \end{align*} and then solve algebraically as I did though.

You're on the right track with the second question, just remember that the period is \displaystyle \begin{align*} 360^{\circ} \end{align*}, which means there is only one period in the region \displaystyle \begin{align*} 0^{\circ} \leq x \leq 360^{\circ} \end{align*}, therefore there must be two solutions in that region. Also remember that the cosine function is positive in the first and FOURTH quadrants.
• Oct 18th 2012, 08:47 PM
skg94
Re: Algebraically solving sin and cos functions, helpl please!
i just have difficulty with the ref and principal angles that is my problem i believe so if i was taught correctly

q=ref=prince
q2 ref+prince=pi
q3=prince-ref=pi
q4=ref+prince=2pi

for the second one

1.04..-30 = -28.95 degrees? saying that its in the first quadrant or the 4th since its negative, is that my reference angle? then i find my princaple angle for the first quadrant?

i now understand the positive and negative sides, for the third one since its negative root3/2 it would make it the 3rd and 4th quadrants so pi/3 and 4pi/3 correct?

thanks for everything so far
• Oct 18th 2012, 09:07 PM
Prove It
Re: Algebraically solving sin and cos functions, helpl please!
Quote:

Originally Posted by skg94
i just have difficulty with the ref and principal angles that is my problem i believe so if i was taught correctly

q=ref=prince
q2 ref+prince=pi
q3=prince-ref=pi
q4=ref+prince=2pi

for the second one

1.04..-30 = -28.95 degrees? saying that its in the first quadrant or the 4th since its negative, is that my reference angle? then i find my princaple angle for the first quadrant?

i now understand the positive and negative sides, for the third one since its negative root3/2 it would make it the 3rd and 4th quadrants so pi/3 and 4pi/3 correct?

thanks for everything so far

It's actually:

Quadrant 1: Angle = Reference Angle.
Quadrant 2: Angle = Pi - Reference Angle
Quadrant 3: Angle = Pi + Reference Angle
Quadrant 4: Angle = 2Pi - Reference Angle.
• Oct 18th 2012, 09:32 PM
skg94
Re: Algebraically solving sin and cos functions, helpl please!
so do i have my -28.95 correct?

so if that is in q4, to find my angle in q1, i do

2pi-(-28.95)?