Solving for t when sin t + cos t = 3/2

[medicalphysicsguy]

Can anyone help me to isolate t when \(\displaystyle sin(t) + cos(t) = \frac{3}{2}\)?

I can use an equivalence to get \(\displaystyle sin(t) + sin(t + \frac{\pi}{2}) = \frac{3}{2}\)

But I don't see where to go next. I take it I can't just take arcsine of both sides!

Thanks, mpg

 

[HallsofIvy]

You could but it wouldn't give you anything useful- arcsin(a+ b) is not equal to arcsin(a)+ arcsin(b).

Instead, go back to the original formula and use the identity sin(a+b)= sin(a)cos(b)+ cos(a)sin(b). In particular, we can think of sin(x)+ cos(x) as sin(x)(1)+ cos(x)(1) and look for b so that cos(b)= sin(b)= 1. Of course, that is impossible because that would give \(\displaystyle cos^2(b)+ sin^2(b)= 2\) when it must equal 1. But we could fix that if we were to multiply (and divide) by \(\displaystyle 1/\sqrt{2}\):
\(\displaystyle \sqrt{2}((1/\sqrt{2})sin(x)+ (1/\sqrt{2})cos(x))\). Now, we have \(\displaystyle sin(b)= cos(b)= 1/\sqrt{2}\) and that gives \(\displaystyle b= \pi/4\) (another possible value is \(\displaystyle \pi- \pi/4= 3\pi/4\)).

That is \(\displaystyle sin(x)+ cos(x)= \sqrt{2}sin(x+ \pi/4)= 3/2\) or \(\displaystyle sin(x)+ cos(x)= \sqrt{2}sin(x+3\pi/4)= 3/2\).

 

[medicalphysicsguy]

That's a great technique, thank you.

 

[Deleted member 4993]

Can anyone help me to isolate t when \(\displaystyle sin(t) + cos(t) = \frac{3}{2}\)?

I can use an equivalence to get \(\displaystyle sin(t) + sin(t + \frac{\pi}{2}) = \frac{3}{2}\)

But I don't see where to go next. I take it I can't just take arcsine of both sides!

Thanks, mpg

General solution for:

\(\displaystyle A*sin(x) + B*cos(x) = C\)

\(\displaystyle \left [\dfrac{A}{\sqrt{A^2+B^2}} *sin(x) + \dfrac{B}{\sqrt{A^2+B^2}}*cos(x)\right ] = \dfrac{C}{\sqrt{A^2+B^2}}\)

If \(\displaystyle \dfrac{C}{\sqrt{A^2+B^2}} \le 1\) then we have solution/s. Let

\(\displaystyle \dfrac{C}{\sqrt{A^2+B^2}} \ = \ cos(\theta)\)

\(\displaystyle \dfrac{A}{\sqrt{A^2+B^2}} \ = \ sin(y)\)

\(\displaystyle \dfrac{B}{\sqrt{A^2+B^2}} \ = \ cos(y)\)

Then

\(\displaystyle \left [\dfrac{A}{\sqrt{A^2+B^2}} *sin(x) + \dfrac{B}{\sqrt{A^2+B^2}}*cos(x)\right ] = \dfrac{C}{\sqrt{A^2+B^2}}\) converts to


\(\displaystyle sin(y)*sin(x) + cos(y)*cos(x) = cos(\theta)\)

\(\displaystyle cos(x+y) = cos(\theta)\)


Solve for 'x' by usual method

 

[Deleted member 4993]

You should see from the above derivation that your problem "does NOT have a solution".

Another way to look at it:

The maximum value of

sin(t) + cos(t) = \(\displaystyle \sqrt{2} < \dfrac{3}{2}\)