# Math Help - need help on 2 simutaneous trig. equations

1. ## need help on 2 simutaneous trig. equations

2 simutaneous equations from my coursework

4=4cosx+3cos(x+y)
3=4sinx+3sin(x+y)

looking for analytical solution, should yield 2 pairs of answers

2. Hi there, you should consider

$4=4\cos(x)+3\cos(x+y)$

$4=4\cos(x)+3(\cos(x)\cos(y)-\sin(x)\sin(y))$

and

$3=4\sin(x)+3sin(x+y)$

$3=4\sin(x)+3(\sin(x)\cos(y)+\cos(x)\sin(y))$

Now expand each expression out

3. ## think algrebra

see soroban's solution

4. Hello, cybernoa!

$\begin{array}{cccc}4\cos x+3\cos(x+y) &=& 4 & {\color{blue}[1]} \\
4\sin x+3\sin(x+y) &=& 3 & {\color{blue}[2]} \end{array}$

Looking for analytical solution, should yield 2 pairs of answers.

$\begin{array}{cccc}
\text{Square {\color{blue}[1]}:} & 16\cos^2x + 24\cos(x+y)\cos x + 9\cos^2(x+y) &=& 16 \\
\text{Square {\color{blue}[2]}:} & 16\sin^2\!x + 24\sin(x+y)\sin x + 9\sin^2(x+y) &=& 9 \end{array}$

$\text{Add: }\;16\underbrace{\bigg[\sin^2x+\cos^2x\bigg]}_{\text{This is 1}} + 24\underbrace{\bigg[\cos(x+y)\cos x + \sin(x+y)\sin x\bigg]}_{\text{This is }\cos[(x+y)-x]}$ $+\; 9\underbrace{\bigg[\sin^2(x+y) + \cos^2(x+y)\bigg]}_{\text{This is 1}} \;=\;25$

We have: . $16 + 24\cos y + 9 \:=\:25 \quad\Rightarrow\quad 24\cos y \:=\:0$

. . . $\cos y \:=\:0 \quad\Rightarrow\quad\boxed{ y \:=\:\frac{\pi}{2},\:\frac{3\pi}{2}}$

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Let $y = \tfrac{\pi}{2}$ in [2]: . $4\sin x + 3\sin\left(x + \tfrac{\pi}{2}\right) \:=\:3$

. . . . $4\sin x + 3\bigg[\sin x\cos\tfrac{\pi}{2} + \cos x\sin\tfrac{\pi}{2}\bigg] \:=\:3$

. . . . . . . . . . $4\sin x + 3\cos x \:=\:3$

Divide by 5: . $\tfrac{4}{5}\sin x + \tfrac{3}{5}\cos x \:=\:\tfrac{3}{5}$ .[3]

Let: . $\cos\theta = \tfrac{4}{5},\;\sin\theta = \frac{3}{5} \quad\Rightarrow\quad \theta = \arcsin\tfrac{3}{5}$ .[4]

Substitute into [3]: . $\cos\theta\sin x + \sin\theta\cos x \:=\:\tfrac{3}{5}$

Then we have: . $\sin(x+\theta) \:=\:\tfrac{3}{5} \quad\Rightarrow\quad x + \theta \:=\:\arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:\arcsin\tfrac{3}{5} - \theta$

Substitute [4]: . $x \:=\:\arcsin\tfrac {3}{5} - \arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:0$

One solution: . $\left(0.\;\tfrac{\pi}{2}\right)$

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Let $y = \frac{3\pi}{2}$ in [2]: . $4\sin x + 3\sin\left(x + \tfrac{3\pi}{2}\right) \:=\:3$

. . . . $4\sin x + 3\bigg[\sin x\cos\tfrac{3\pi}{2} + \cos x\sin\tfrac{3\pi}{2}\bigg] \:=\:3$

. . . . . . . . . . $4\sin x - 3\cos x \:=\:3$

Divide by 5: . $\tfrac{4}{5}\sin x + \tfrac{3}{5}\cos x \:=\:\tfrac{3}{5}$ .[5]

Let: . $\cos\theta = \tfrac{4}{5},\;\sin\theta = \tfrac{3}{5} \quad\Rightarrow\quad \theta \,=\,\arcsin\tfrac{3}{5}$ .[6]

Substitute into [5]: . $\cos\theta\sin x - \sin\theta\cos x \:=\:\tfrac{3}{5}$

Then we have: . $\sin\left(x - \theta\right) \:=\:\tfrac{3}{5} \quad\Rightarrow\quad x-\theta \:=\:\arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:\arcsin\tfrac{3}{5} + \theta$

Substitute [6]: . $x \:=\:\arcsin\tfrac{3}{5} + \arcsin\tfrac{3}{5} \quad\Rightarrow\quad x \:=\:2\arcsin\tfrac{3}{5}$

Another solution: . $\left(2\arcsin\tfrac{3}{5},\;\tfrac{3\pi}{2}\right )$

5. Originally Posted by Soroban
Hello, cybernoa!
Soroban, I've said it once, and I'll say it again. You're presentation is impeccable.