Question

Evaluate cos 48 cos 42- sin 48 sin 42

Answer 1

$\cos48 \cos42 - \sin48 \sin42$

As we know that :

$\cos( \alpha + \beta ) = cos \alpha cos \beta - \sin \alpha \sin \beta$

$here \: \alpha = 48 \: and \: \beta = 42 \\ \\ = > \cos(48 + 42) \\ \\ = > \cos90 = 0$

So, the answer will be 0

Answer 2

Answer:

$\cos48^\circ\cos42^\circ-\sin48^\circ\sin42^\circ=0$

Step-by-step explanation:

To evaluate: $\cos48^\circ\cos42^\circ-\sin48^\circ\sin42^\circ$

Formula: cos A cos B - sin A sin B = cos(A+B)

where, A = 48°, B = 42°

$\Rightarrow \cos48^\circ\cos42^\circ-\sin48^\circ\sin42^\circ$

$\Rightarrow \cos(48^\circ+42^\circ)$

$\Rightarrow \cos(90^\circ)$

$\Rightarrow 0$

Hence, the value of $\cos48^\circ\cos42^\circ-\sin48^\circ\sin42^\circ=0$