+4 $$\underset{\,\,\,\,{\textcolor[rgb]{0.66,0.66,0.66}{\rightarrow {\mathtt{sin0, cos}}}}}{{solve}}{\left(\begin{array}{l}{\mathtt{Sin0}}={{\mathtt{5}}}^{{\mathtt{7}}}\\
{cos}\left(=\mathrm{ERROR}\end{array}\right)} \Rightarrow \left\{ \begin{array}{l}{\mathtt{sin0}} = {\mathtt{78\,125}}\\
{cos}\left( = {\mathtt{ERROR}}\\
\end{array} \right\}$$What is csc if sin =5/7
+118677 Thanks Chris, the trig identity works fine - Here's another take on the problem.
ok sine is postitive for theta must be in the first or second quadrant. This just means that Cosθ can be positive or negative. To find the one in the first quad just draw a right angled triange. Put your info on it and find the third side using Pythagoras' Theorems.
Like this. So reading straight from the triangle $$cos\theta=\pm \frac{2\sqrt6}{7}$$
+129852 Using a trig identity, we have
1 - (5/7)^2 = (cosx)^2
1 - 25/49 = (cosx)^2
24/49 = (cosx)^2 take the square root of both sides, we have
±√(24)/7 = ±[2√(6)]/7 = cosx
This makes sense...the sine is positive in the first two quadrants, so the cos is positive in the first, but negative ni the second.......
+118677 Thanks Chris, the trig identity works fine - Here's another take on the problem.
ok sine is postitive for theta must be in the first or second quadrant. This just means that Cosθ can be positive or negative. To find the one in the first quad just draw a right angled triange. Put your info on it and find the third side using Pythagoras' Theorems.
Like this. So reading straight from the triangle $$cos\theta=\pm \frac{2\sqrt6}{7}$$
