Question

Why is $\cos \left(90°-x\right)=\sin \left(x\right)$ and $\sin \left(90°-x\right)=\cos \left(x\right)$?

Answer 1

Prove that $\cos \left(90°-x\right)=\sin \left(x\right)$ and $\sin \left(90°-x\right)=\cos \left(x\right)$.

Apply sum and difference formulas:

$$\begin{gathered} \cos \left(a-b\right)=\cos \left(a\right)\cos \left(b\right)+\sin \left(a\right)\sin \left(b\right)...\left(1\right) \\ \sin \left(a-b\right)=\sin \left(a\right)\cos \left(b\right)-\cos \left(a\right)\sin \left(b\right)...\left(2\right) \end{gathered}$$

Using the first equation where $a=90°$ and $b=x$ we have,

$\begin{array}{rcl}\cos \left(90°-x\right)& =& \cos \left(90°\right)\cos \left(x\right)+\sin \left(90°\right)\sin \left(x\right)\\ & \Rightarrow & \cos \left(90°-x\right)=0\cos \left(x\right)+1\sin \left(x\right)\left\{\cos \left(90°\right)=0,\sin \left(90°\right)=1\right\}\\ \therefore \cos \left(90°-x\right)& =& \sin \left(x\right)\end{array}$

Using the second equation where $a=90°$ and $b=x$ we have,

$\begin{array}{rcl}\sin \left(90°-x\right)& =& \sin \left(90°\right)\cos \left(x\right)-\cos \left(90°\right)\sin \left(x\right)\\ & \Rightarrow & \sin \left(90°-x\right)=1\cos \left(x\right)-0\sin \left(x\right)\left\{\sin \left(90°\right)=1,\cos \left(90°\right)=0\right\}\\ \therefore \sin \left(90°-x\right)& =& \cos \left(x\right)\end{array}$

Hence, $\cos \left(90°-x\right)=\sin \left(x\right)$ and $\sin \left(90°-x\right)=\cos \left(x\right)$.