Can You Determine the Value of A When B Equals 4 in a Trigonometric Equation?

[NotaMathPerson]

A=3sinx+4cosx and B=3cosx-4sinx if B = 4 find A.

What i tried is to use 4=3cosx-4sinx and solve for cosx

now cosx = (4+4sinx)/3 plug this into A

I end up getting A = (25sinx+16)/3 am I correct?

 

[soroban]

Re: Trigonometric equatio

$$A\;=\,3\sin x+4\cos x\,\text{ and }\,B\,=\,3\cos x-4\sin x.$$

$$\text{If }B = 4.\,\text{ find }A.$$

$$\text{If }B = 4,\,\text{then }\,x = \tfrac{3\pi}{2}$$

$$\text{Then: }\,A \:=\:3\sin\tfrac{3\pi}{2} + 4\cos\tfrac{3\pi}{2} \:=\:3(-1) + 4(0) \:=\: -3$$

 

[NotaMathPerson]

Re: Trigonometric equatio

$$\text{If }B = 4,\,\text{then }\,x = \tfrac{3\pi}{2}$$

$$\text{Then: }\,A \:=\:3\sin\tfrac{3\pi}{2} + 4\cos\tfrac{3\pi}{2} \:=\:3(-1) + 4(0) \:=\: -3$$

Hello soroban!

How did you get the value for x?

 

[soroban]

Re: Trigonometric equatio

I wrote: If $$B = 4,\,$$ then $$x = \tfrac{3\pi}{2}$$
Here is the reasoning behind that claim.

We are given: $$B \,=\,3\cos x - 4\sin x$$
. . And we are told that: $$B = 4.$$
That is: $$3\cos x - 4\sin x \:=\:4$$

This is true if $$\cos x = 0$$ and $$\sin x = -1.$$
Therefore: $$x \,=\,\tfrac{3\pi}{2}$$

 

[Prove It]

Re: Trigonometric equatio

I wrote: If $$B = 4,\,$$ then $$x = \tfrac{3\pi}{2}$$
Here is the reasoning behind that claim.

We are given: $$B \,=\,3\cos x - 4\sin x$$
. . And we are told that: $$B = 4.$$
That is: $$3\cos x - 4\sin x \:=\:4$$

This is true if $$\cos x = 0$$ and $$\sin x = -1.$$
Therefore: $$x \,=\,\tfrac{3\pi}{2}$$

This isn't very rigorous, although I am impressed by your intuition :)

I would be more inclined to try to solve the problem directly...

$\displaystyle \begin{align*} A &= 3\sin{(x)} + 4\cos{(x)} \\ B &= 3\cos{(x)} - 4\sin{(x)} \\ \\ A^2 &= \left[ 3\sin{(x)} + 4\cos{(x)} \right] ^2 \\ B^2 &= \left[ 3\cos{(x)} - 4\sin{(x)} \right] ^2 \\ \\ A^2 &= 9\sin^2{(x)} + 24\sin{(x)}\cos{(x)} + 16\cos^2{(x)} \\ B^2 &= 9\cos^2{(x)} - 24\sin{(x)}\cos{(x)} + 16\sin^2{(x)} \\ \\ A^2 + B^2 &= 9\sin^2{(x)} + 24\sin{(x)}\cos{(x)} + 16\cos^2{(x)} + 9\cos^2{(x)} - 24\sin{(x)}\cos{(x)} + 16\sin^2{(x)} \\ A^2 + B^2 &= 25\left[ \sin^2{(x)} + \cos^2{(x)} \right] \\ A^2 + B^2 &= 25 \\ A^2 + 4^2 &= 25 \\ A^2 + 16 &= 25 \\ A^2 &= 9 \\ A &= \pm 3 \end{align*}$

Now you just have to check for extraneous solutions, as you have had to square the equations to be able to solve them.