Question

Please help with these two questions I'm very confused!

Answer 1

1)

$\bf \textit{Cofunction Identities} \\\\ sin\left(\frac{\pi}{2}-\theta\right)=cos(\theta) \qquad cos\left(\frac{\pi}{2}-\theta\right)=sin(\theta) \\\\\\ \textit{also recall that }sin(-\theta )=-sin(\theta )\\\\ -------------------------------$

$\bf sin\left( \theta -\frac{\pi }{2} \right)\implies sin\left[-\left( \frac{\pi }{2}-\theta \right) \right]\implies -sin\left( \theta -\frac{\pi }{2} \right)\implies -cos(\theta ) \\\\\\ \textit{and since }cos(\theta )=0.54\qquad then\qquad -cos(\theta )\implies -0.54$



2)

$\bf \textit{Cofunction Identities} \\\\ sin\left(\frac{\pi}{2}-\theta\right)=cos(\theta) \qquad cos\left(\frac{\pi}{2}-\theta\right)=sin(\theta) \\\\\\ tan\left(\frac{\pi}{2}-\theta\right)=cot(\theta)\qquad cot\left(\frac{\pi}{2}-\theta\right)=tan(\theta) \\\\\\ \textit{also recall }sin(-\theta )=-sin(\theta )\qquad cos(-\theta )=cos(\theta )\\\\ -------------------------------$

$\bf tan\left( x-\frac{\pi }{2} \right)\implies \cfrac{sin\left( x-\frac{\pi }{2} \right)}{cos\left( x-\frac{\pi }{2} \right)}\implies \cfrac{sin\left[ -\left( \frac{\pi }{2}-x \right) \right]}{cos\left[ -\left( \frac{\pi }{2}-x \right) \right]} \\\\\\ \cfrac{-sin\left( \frac{\pi }{2}-x \right)}{cos\left( \frac{\pi }{2}-x \right)}\implies -tan\left( \frac{\pi }{2}-x \right)\implies -cot(x) \\\\\\ \textit{and since }cot(x)=-0.18\qquad then\qquad -cot(x)\implies 0.18$