Solving for the dependant variable in sinusoidal functions .

[hannahsiao]

I have a word problem with the equation being:

h=1.55cos[360/12(t-10.75)]+2.7

Then it says let h=3.2, how do I solve this?! All I can do is take away the 2.7 and put it on the other side, but then what?

 

[HallsofIvy]

Yes, setting h= 3.2 and subtracting 2.7 from both sides give
.5= 1.55 cos(360/12(t- 10.75))

Now, divide both sides by 1.55 to get
0.3226= cos(360/12(t- 10.75))

You could probably see that for yourself. Now, to get rid of any function on one side of an equation, use its "inverse" function. For cosine, that is the "arccosine" or $cos^{-1}(x)$ (NOT reciprocal!)

cos^{-1}(.3226)= 360/12(t- 10.75)

Now, I have some questions my self. Is that "angle" in degrees or radians? Normally in equations involving sine or cosine, it is radians but that "360" makes me wonder. Also is that (t- 10.75) in the denominator of numerator? Is it
$$\frac{360}{12(t- 10.75)}$$
or
$$\frac{360}{12}(t- 10.75)$$?

(In either case 360/12= 30.)

Finally, are you looking for a single solution, solutions in a given range, or all solutions?
Cosine is periodic with period $2\pi$ (or 360 degrees) so adding any multiple of $2\pi$ gives another solution. Also $cos(2\pi- x)= cos(x)$ so we have two solutions within each [itex]2\pi[/tex] interval.

 

[hannahsiao]

(t-10.75) is the second option beside 360/12 or 30.

 

[hannahsiao]

so no 71(rounded)=[30(t-10.75)]
do i now divide by 30 ? and then add 10.75 to find t?

 

[HallsofIvy]

so no 71(rounded)=[30(t-10.75)]
do i now divide by 30 ? and then add 10.75 to find t?

IF the argument lf cosine is in degrees and IF your angle is between 0 and 90 degrees, then yes.