Why does cos(90 - x) = sin(x) and sin(90 - x) = cos(x)?

Apr 17, 2015

Note that the image below is only for $x$ in Q1 (the first quadrant).
If you wish you should be able to draw it with $x$ in any quadrant.

Definition of $\sin \left(x\right)$

(side opposite angle x)//(hypotenuse)

Definition of $\cos \left({90}^{\circ} - x\right)$

(side adjacent to angle (90^@-x))//(hypotenuse)

but (side opposite angle x) = (side adjacent to angle $\left({90}^{\circ} - x\right)$

Therefore

$\sin \left(x\right) = \cos \left({90}^{\circ} - x\right)$

Similarly

$\cos \left(x\right) = \sin \left({90}^{\circ} - x\right)$

Feb 7, 2016

These can also be proven using the sine and cosine angle subtraction formulas:

$\cos \left(\alpha - \beta\right) = \cos \left(\alpha\right) \cos \left(\beta\right) + \sin \left(\alpha\right) \sin \left(\beta\right)$

$\sin \left(\alpha - \beta\right) = \sin \left(\alpha\right) \cos \left(\beta\right) - \cos \left(\alpha\right) \sin \left(\beta\right)$

Applying the former equation to $\cos \left({90}^{\circ} - x\right)$, we see that

$\cos \left({90}^{\circ} - x\right) = \cos \left({90}^{\circ}\right) \cos \left(x\right) + \sin \left({90}^{\circ}\right) \sin \left(x\right)$

$\cos \left({90}^{\circ} - x\right) = 0 \cdot \cos \left(x\right) + 1 \cdot \sin \left(x\right)$

$\cos \left({90}^{\circ} - x\right) = \sin \left(x\right)$

Applying the latter to $\sin \left({90}^{\circ} - x\right)$, we can also prove that

$\sin \left({90}^{\circ} - x\right) = \sin \left({90}^{\circ}\right) \cos \left(x\right) - \cos \left({90}^{\circ}\right) \sin \left(x\right)$

$\sin \left({90}^{\circ} - x\right) = 1 \cdot \cos \left(x\right) - 0 \cdot \sin \left(x\right)$

$\sin \left({90}^{\circ} - x\right) = \cos \left(x\right)$