The relationship between shear stress $\sigma_y$ and direct stress $\sigma_x$ and $\sigma_y $ was found to be of the form:
$$\tag{1} \sigma = \frac{\sigma_x - \sigma_y}{2}~ \sin e (2\theta) $$
Find the value of the direct stress $\sigma_y$ when the value of the shear stress $\sigma$ is $25~\text{MNm}^{-2}$, the value of $\sigma_x$ is $80~\text{MNm}^{-2}$ and the value of the angle of the plane $\theta$ to the horizontal is $25^{\circ}$
Looking for help with this.
Would I use a hyperbolic function for this?
$$ \sinh x = \frac{e^x - e^{-x}}{2} $$
I think as fGDu94 said, it's just a matter of rearranging the equation.
leads to
$$ \sigma_y =\sigma_x-\frac{(2\sigma)}{\sin e (2\theta)} $$
By replacing with the given values we obtain:
$$ \sigma_y = 80 - \frac{(2*25)}{\sin e (2*25deg)}\\= 80 - \frac{(2*25)}{\sin e (2*\frac{25*Pi}{180}rad)} \\= 80-71.8663237517.. \\=8.13367624827.. \approx 8 $$
NB: if the exponential function in the sine function happens to be a typo, the result would then become:
$$ \sigma_y =\sigma_x-\frac{(2\sigma)}{\sin (2\theta)} \\= 80- 65.2703644666.. = 14.7296355334..\approx 15$$