# Thread: Simple Trigonometry help, and logs

1. ## Simple Trigonometry help, and logs

How would i go about working out:

1. cos(x) + sin(x) = (1/3), -pi < x < pi

2. ln(y+1) - ln(y) = 0.85

2. Hello,

For the first one, I'd make a substitution with u=cos(x).

As cos²(x)+sin²(x)=1, sin(x)=sqrt(1-cos²(x))=sqrt(1-u²)

Thus we have $u+\sqrt{1-u^2}=1/3$

$1/3-u=\sqrt{1-u^2}$

Square the whole equation and solve for u, then deduct the values for x.

For the second one : $\ln(a)-\ln(b)=\ln(\frac ab)$
So it's equivalent to :

$\frac{y+1}{y}=e^{0.85}$

3. Hello, Stylis10!

Another approach to #1 . . .

$1.\;\;\cos x + \sin x \:=\:\frac{1}{3}\qquad -\pi < x < \pi$
Divide by $\sqrt{2}\!:\qquad \frac{1}{\sqrt{2}}\cos x + \frac{1}{\sqrt{2}}\sin x \:=\:\frac{1}{3\sqrt{2}}$

And we have: . $\sin\frac{\pi}{4}\cos x + \cos\frac{\pi}{4}\sin x \:=\:\frac{\sqrt{2}}{6}$

. . . . . . . . . . . . . . . . . $\sin\left(\frac{\pi}{4}+x\right) \:=\:\frac{\sqrt{2}}{6}$

Hence: . $\frac{\pi}{4}+x \:=\:\sin^{-1}\left(\frac{\sqrt{2}}{6}\right) \quad\Rightarrow\quad x \;=\;\sin^{-1}\left(\frac{\sqrt{2}}{6}\right) - \frac{\pi}{4}$

. . Therefore: . $x \;=\;\begin{Bmatrix}\text{-}0.547457039 \\ \;2.118253365 \end{Bmatrix}$