Question & Answer

Find the value of \[\cos \left( 270{}^\circ +\theta \right)\cos \left( 90{}^\circ +\theta \right)-\cos \left( 270{}^\circ +\theta \right)\cos \theta .\]

ANSWER Verified Verified
Hint: Compute values of $\cos \left( 90{}^\circ +\theta \right)$and $\cos \left( 270{}^\circ +\theta \right)$using the formula;
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
Then put these values in the expression and compute the answer.

Complete step-by-step answer:
We have to find the values of \[\cos \left( 270{}^\circ +\theta \right)\cos \left( 90{}^\circ +\theta \right)-\cos \left( 270{}^\circ +\theta \right)\cos \theta .\]
For solving the expression, we need to change each term into terms of trigonometric ratios of $\theta \left( i.e.\ \sin \theta \ or\ \cos \theta \right)$.
Using the formula $\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$,
$\cos \left( 90+\theta \right)=\cos 90{}^\circ \times \cos \theta -\sin 90{}^\circ \times \sin \theta $
We know $\cos 90{}^\circ =0\ and\ \sin 90{}^\circ =1$
So, $\cos \left( 90+\theta \right)=0-\sin \theta $
$\Rightarrow \cos \left( 90+\theta \right)=-\sin \theta $
Similarly, $\cos \left( 270+\theta \right)=\cos 270\times \cos \theta -\sin 270\times \sin \theta $
We know that $\cos 270{}^\circ =0\ and\ \sin 270{}^\circ =-1$
So, $\cos \left( 270+\theta \right)=0-\left( -1 \right)\times \sin \theta $
$\Rightarrow \cos \left( 270{}^\circ +\theta \right)=\sin \theta $
Now putting these values in the given expression we will get;
  & \cos \left( 270+\theta \right)\cos \left( 90+\theta \right)-\cos \left( 270+\theta \right)\cos \theta \\
 & =\sin \theta \times \left( -\sin \theta \right)-\left( \sin \theta \right)\cos \theta \\
 & ={{\sin }^{2}}\theta -\sin \theta \cos \theta \\
Taking $-\sin \theta $ common, we will get;
$=-\sin \theta \left( \sin \theta +\cos \theta \right)$
Hence, $\cos \left( 270{}^\circ +\theta \right)\cos \left( 90{}^\circ +\theta \right)-\cos \left( 270{}^\circ +\theta \right)\cos \theta =-\sin \theta \left( \cos \theta +\sin \theta \right)$

Note: We should memorize the values of $\cos \left( 90{}^\circ \pm \theta \right),\cos \left( 180{}^\circ \pm \theta \right),\cos \left( 270{}^\circ \pm \theta \right),\sin \left( 90{}^\circ \pm \theta \right),\sin \left( 180{}^\circ \pm \theta \right),\sin \left( 270{}^\circ \pm \theta \right),$
$\tan \left( 90{}^\circ \pm \theta \right),\tan \left( 180{}^\circ \pm \theta \right),\tan \left( 270{}^\circ \pm \theta \right)$ and so on. As these values are commonly used.
Easy trick for remembering these values;
  & \cos \left( K\pm \theta \right)=\left( \pm \right)\sin \theta \ if\ K=90{}^\circ ,270{}^\circ ,450{}^\circ ,..... \\
 & =\left( \pm \right)\cos \theta \ if\ K=180{}^\circ ,360{}^\circ ,540{}^\circ ,..... \\
 & \sin \left( K\pm \theta \right)=\left( \pm \right)\sin \theta \ if\ K=180{}^\circ ,360{}^\circ ,540{}^\circ ,..... \\
 & =\left( \pm \right)\cos \theta \ if\ K=90{}^\circ ,180{}^\circ ,450{}^\circ ,..... \\
Sign is decided by checking that in which quadrant angle will line.
If the angel is in;
1st quadrant $\to $all trigonometric ratios will be positive
2nd quadrant $\to \sin \theta \ and\ \cos ec\theta $ will be positive only
3rd quadrant $\to \tan \theta \ and\ \cot \theta $ will be positive only
4th quadrant $\to \cos \theta \ and\ \sec \theta $ will be positive only.
Mnemonics used for memorizing this\[\to \begin{matrix}
   \underset{{{1}^{st}}\ quadrant}{\mathop{\underline{ALL}}}\, & \underset{{{2}^{nd}}\ quadrant}{\mathop{\underline{SIN}}}\, & \underset{{{3}^{rd}}\ quadrant}{\mathop{\underline{TAN}}}\, & \underset{{{4}^{th}}\ quadrant}{\mathop{\underline{COS}}}\, \\