Given cos(pi/2-x)=3/5, cosx=4/5 to find the remaining trigonometric function?

Jan 5, 2017

To find the value of the sine, use the identity for the cosine of the difference of two angles:
$\cos \left(A - B\right) = \cos \left(A\right) \cos \left(B\right) + \sin \left(A\right) \sin \left(B\right)$. Please see below.

Explanation:

The identity for the cosine of the difference of two angles is:

$\cos \left(A - B\right) = \cos \left(A\right) \cos \left(B\right) + \sin \left(A\right) \sin \left(B\right)$

Given: $A = \frac{\pi}{2} \mathmr{and} B = x$

$\cos \left(\frac{\pi}{2} - x\right) = \cos \left(\frac{\pi}{2}\right) \cos \left(x\right) + \sin \left(\frac{\pi}{2}\right) \sin \left(x\right)$

The first term on the right disappears, because $\cos \left(\frac{\pi}{2}\right) = 0$, and the second term on the right becomes $\sin \left(x\right)$, because $\sin \left(\frac{\pi}{2}\right) = 1$:

$\cos \left(\frac{\pi}{2} - x\right) = \sin \left(x\right) = \frac{3}{5}$

$\sin \left(x\right) = \frac{3}{5}$

$\tan \left(x\right) = \sin \frac{x}{\cos} \left(x\right) = \frac{\frac{3}{5}}{\frac{4}{5}}$

$\tan \left(x\right) = \frac{3}{4}$

$\cot \left(x\right) = \frac{1}{\tan} \left(x\right)$

$\cot \left(x\right) = \frac{4}{3}$

$\csc \left(x\right) = \frac{1}{\sin} \left(x\right)$

$\csc \left(x\right) = \frac{5}{3}$

$\sec \left(x\right) = \frac{1}{\cos} \left(x\right)$

$\sec \left(x\right) = \frac{5}{4}$