Question

cos 78×cos 12-sin 27 ×cos3=

Answer 1

I guess the question is wrong.
If it is correct then the answer is
sin 12× cos 12 - sin 27×cos 3

Answer 2

0.65 or $\frac{13}{20}$

Using sine and cosine formulas of:

(i) cos a × cos b = cos (a-b) + cos (a+b) / 2

(ii) sin a × sin b = sin (a+b) + sin (a-b) / 2

Therefore here,

cos a = cos 78 and cos b = cos 12

and sin a = 27 and sin b = sin 3.

Substituting,

cos a × cos b - (sin a × sin b) = cos (a-b) + cos (a+b) / 2 - ( sin (a+b) + sin (a-b) / 2)

∴ cos 78 × cos 12 - sin 27 × cos 3 = cos (78-12) + cos (78+12) / 2 - (sin (27+3) + sin (27-3) / 2)

∴ cos 78 × cos 12 - sin 27 × cos 3 = cos 66 + cos 90 / 2 - (sin 30 + sin 24 / 2)

∴ cos 78 × cos 12 - sin 27 × cos 3 = cos 66 / 2 - ( 1 - 2*sin 24 / 4)

∴ cos 78 × cos 12 - sin 27 × cos 3 = 0.6567