1. ## please check trig work2

1. in the general equation $\displaystyle y= acos kx$ the period is determined by the value of x

2. in the general equation $\displaystyle y= acos k(x+b)$ the phase shift is determined by the value of b

3. the period of the function $\displaystyle y= 3cos \frac{2}{3} x-4$ is -4

4. if \theta is in standard position, and 0 <or equal $\displaystyle \theta$ <or equal $\displaystyle 2\pi$, if $\displaystyle sin\theta=\frac{3}{5}$ the value of tan $\displaystyle \theta$ is $\displaystyle \frac{4}{3}$

5. the equation of a sine function with amplitude 4 and period of $\displaystyle 90^o$ is $\displaystyle y=4sin90x$

2. Originally Posted by william
1. in the general equation $\displaystyle y= acos kx$ the period is determined by the value of x
no, period is $\displaystyle \frac{2\times \pi}{k}$

Originally Posted by william

2. in the general equation $\displaystyle y= acos k(x+b)$ the phase shift is determined by the value of b
I agree

Originally Posted by william

3. the period of the function $\displaystyle y= 3cos \frac{2}{3} x-4$ is -4
refer to my response to question 1.

Originally Posted by william
1

4. if \theta is in standard position, and 0 <or equal $\displaystyle \theta$ <or equal $\displaystyle 2\pi$, if $\displaystyle sin\theta=\frac{3}{5}$ the value of tan $\displaystyle \theta$ is $\displaystyle \frac{4}{3}$
as $\displaystyle tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$

then $\displaystyle tan(\theta)= \frac{3}{4} \neq \frac{4}{3}$

Originally Posted by william
5. the equation of a sine function with amplitude 4 and period of $\displaystyle 90^o$ is $\displaystyle y=4sin90x$
period is $\displaystyle \frac{2\times \pi}{k}$

therefore $\displaystyle \frac{\pi}{2} = \frac{2\times \pi}{k}$

this means $\displaystyle k= 4$ and $\displaystyle y=4\sin(4x)$

3. Originally Posted by pickslides
no, period is $\displaystyle \frac{2\times \pi}{k}$

I agree

refer to my response to question 1.

as $\displaystyle tan(\theta) = \frac{sin(\theta)}{cos(\theta)}$

then $\displaystyle tan(\theta)= \frac{3}{4} \neq \frac{4}{3}$

period is $\displaystyle \frac{2\times \pi}{k}$

therefore $\displaystyle \frac{\pi}{2} = \frac{2\times \pi}{k}$

this means $\displaystyle k= 4$ and $\displaystyle y=4\sin(4x)$
thanks i appreciate the help,my updated answers:

1. can it be just k? can you just say k?

3. 2/3 pi

are they good now?

4. no, it can't just be k.

Q1

The period is found as $\displaystyle \frac{2\pi}{k}$ therefore this is the general solution.

Q3

$\displaystyle \frac{2\pi}{k}$ therefore youe answer is $\displaystyle \frac{2\pi}{\frac{2}{3}} = 3\pi$ so the period or length of the cosine wave is $\displaystyle 3\pi$