Help with trig q ?

[Oors]

help is much appreciated

 

[MarkFL]

I would use a linear combination identity to write:

[MATH]3\cos(15t)+4\sin(15t)=5\sin\left(15t+\arctan\left(\frac{3}{4}\right)\right)[/MATH]
Now, consider:

[MATH]-1\le\sin(\theta)\le1[/MATH]
[MATH]-5\le5\sin(\theta)\le5[/MATH]
[MATH]5\le10+5\sin(\theta)\le15[/MATH]
Can you proceed?

 

[Oors]

I would use a linear combination identity to write:

[MATH]3\cos(15t)+4\sin(15t)=5\sin\left(15t+\arctan\left(\frac{3}{4}\right)\right)[/MATH]
Now, consider:

[MATH]-1\le\sin(\theta)\le1[/MATH]
[MATH]-5\le5\sin(\theta)\le5[/MATH]
[MATH]5\le10+5\sin(\theta)\le15[/MATH]
Can you proceed?

so I worked out for part A value for theta is 53.13 degrees and the maximum is at R= 5 however I just don't know how to proceed with part B and c

 

[MarkFL]

Yes, I agree that the maximum value of the sum of the two trig functions is 5, and this maximum occurs for:

[MATH]15t+\arctan\left(\frac{3}{4}\right)=\frac{\pi}{2}[/MATH]
[MATH]t=\frac{1}{15}\left(\frac{\pi}{2}-\arctan\left(\frac{3}{4}\right)\right)\approx0.0618196812[/MATH]
However, if we are to use the suggested form, we need to write:

[MATH]5\sin\left(15t+\arctan\left(\frac{3}{4}\right)\right)=5\cos\left(15t-\left(\frac{\pi}{2}-\arctan\left(\frac{3}{4}\right)\right)\right)[/MATH]
And of course we get the same value for \(t\) at the maximum.

Use this to find the minimum temperature:

[MATH]5\le10+5\sin(\theta)\le15[/MATH]
And to find when the minimum occurs, use either form where:

[MATH]\cos(\pi)=\sin\left(\frac{3\pi}{2}\right)=-1[/MATH]