Question

Prove that $\text{cos}\theta +\text{sin}({270}^{\circ }+\theta )-\text{sin}({270}^{\circ }-\theta )+\text{cos}({180}^{\circ }+\theta )$

Answer 1

= $cos\theta +sin({270}^{\circ }+\theta )-sin({270}^{\circ }-\theta )+cos({180}^{\circ }+\theta )$

= $cos\theta +sin[{180}^{\circ }+({90}^{\circ }+\theta \left)\right]-sin[{180}^{\circ }+({90}^{\circ }-\theta \left)\right]+cos[{180}^{\circ }+\theta ]$

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

$[\because \text{sin}({180}^{\circ }+x)=-\text{sin}x,\text{cos}({180}^{\circ }+x)=-cosx]$

= $\text{cos}\theta -\text{cos}\theta +\text{cos}\theta -\text{cos}\theta =0$

= $[\because \text{sin}({90}^{\circ }+\theta )=\text{cos}\theta ,\text{sin}({90}^{\circ }-\theta \left)\right]$