Algebraically Solving Trig Functions for T: Tips and Tricks

[gjb19]

Ok, we'll basically we have two different functions and we're told to set them equal and then algebracially solve them for T.

Here are the functions:

y = 20 sin(pie/15 t) + 25

y = -10cos(2pie/15 t) + 12

It gives a hint and says to use a double-angle forumla and the quadratic forumla to help if needed.

Does anyone have any idea on where to start? I'm totally confused.

Thanks a lot!

 

[acm]

y = -10Cos(2pi/15 t) + 12
y = -10({Cos(pi/15t)}^2 - {Sin(pi/15t)}^2) + 12

From here it is pretty obvious what to do, get the expression in terms of Sin and solve for Y.

 

[Architeuthis Dux]

I would suggest starting by reviewing the basic properties and identities of trigonometric functions. This will help in understanding the given functions and how to solve them for T. Additionally, using a graphing calculator or software can also be helpful in visualizing the functions and their intersections.

One approach to solving these types of equations is to use substitution. In this case, we can substitute the first equation into the second equation, replacing y with the given expression. This will result in an equation with only one variable, t. From there, we can use the quadratic formula to solve for t and then plug that value back into the original equation to find the corresponding value of y.

Another approach is to use the double-angle formula for cosine, which states that cos(2x) = 1 - 2sin^2(x). By rearranging this equation, we can express sin^2(x) in terms of cos(2x) and then use it to substitute for sin(pie/15 t) in the first equation. This will again result in an equation with only one variable, t, which can be solved using algebraic techniques.

It is also important to keep in mind the domain and range of the trigonometric functions involved. In this case, the domain of both functions is all real numbers, while the range of the first function is y≥ -5 and the range of the second function is y≤ 22. Therefore, when solving for t, we must consider these restrictions and choose the appropriate solution.

In conclusion, solving trigonometric functions algebraically for T may seem challenging at first, but with a solid understanding of trigonometric properties and identities, along with the use of substitution and the quadratic formula, it can be solved systematically and accurately.