Question
Evaluate the following:
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
✓
Answer
We have to find the value of the following expression
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ\dots(1)$
$\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2},\sin60^\circ=\frac{\sqrt3}{2}\cos60^\circ=\frac{1}{2}$
Now
So by substituting above values in equation (1)
We get,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
$=\frac{1}{2}\times\frac{1}{\sqrt2}-\frac{\sqrt3}{2}\times\frac{1}{\sqrt2}$
$=\frac{1}{2\sqrt2}-\frac{\sqrt3}{2\sqrt2}$
$=\frac{1-\sqrt3}{2\sqrt2}$
Therefore,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ=\frac{1-\sqrt3}{2\sqrt2}$
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ\dots(1)$
$\sin45^\circ=\cos45^\circ=\frac{1}{\sqrt2},\sin60^\circ=\frac{\sqrt3}{2}\cos60^\circ=\frac{1}{2}$
Now
So by substituting above values in equation (1)
We get,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ$
$=\frac{1}{2}\times\frac{1}{\sqrt2}-\frac{\sqrt3}{2}\times\frac{1}{\sqrt2}$
$=\frac{1}{2\sqrt2}-\frac{\sqrt3}{2\sqrt2}$
$=\frac{1-\sqrt3}{2\sqrt2}$
Therefore,
$\cos60^\circ\cos45^\circ-\sin60^\circ\sin45^\circ=\frac{1-\sqrt3}{2\sqrt2}$
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(i) Let A be the event of getting a heart card.
There are ⬜ cards of heart. $\quad$$\quad$∴n(A) = ⬜
$P(A)=\frac{n(A)}{n(S)}$
$\therefore P(A)$= $[-]$
(ii) Let B be the event of not getting a face card.
There are ⬜ cards which are not face cards.$\quad$$\quad$∴n(B) = ⬜
$P(B)=\frac{n(B)}{n(S)}$
$\therefore P(B)$= $[-]$