Question

The value of cos²10º - cos10º cos50º + cos² 50º is:
(A) 3/2(1 + cos20º)
(B) 3/4
(C) 3/2
(D) (3/4) + cos20º

Answer 1

answer : option (B) 3/4

given, cos²10° - cos10°. cos50° + cos²50°

= (cos10° + cos50°)² - 2cos10°.cos50° -cos10°.cos50°

we know, cosC + cosD = 2cos(C + D)/2. cos(C - D)/2

= (2cos30°.cos20°)² - 3cos10°.cos50°

= (2 × √3/2 .cos20°)² - 3/2 [2cos10°.cos50°]

= 3cos²20° - 3/2 [2cos10°.cos50°]

we know, 2cosC. cosD = cos(C + D) + cos(C - D)

= 3cos²20° - 3/2 [cos60° + cos40°]

= 3cos²20° - 3/2 × 1/2 - 3/2 cos40°

we know, cos2x = 2cos²x - 1

= 3cos²20° - 3/4 - 3/2 [2cos²20° - 1]

= 3cos²20° - 3/4 - 3cos²20° + 3/2

= 3/2 - 3/4

= 3/4

Answer 2

Answer:

3/4

Step-by-step explanation:

The re-arrangement of the equation can be done this way:

$cos^{2} 10 + cos^{2} 50 - cos 10cos50$

Then the solution for the question goes like this

$cos^{2} 10 + cos^{2} 50 - cos 10cos50\\=cos^{2} 10 + 1-sin^{2}50 - cos10cos50$                                        (∵ $cos^{2} A + sin^{2} A=1$)

$=cos(10+50)cos(10-50) + 1 -cos10cos50$

$=cos60cos40 + 1 -(2(cos10cos50))/2$                                     (∵$cos(-\alpha ) = cos(\alpha )$)

$=(cos40)/2 +1-((cos(10+50))/2+(cos(10-50))/2)$

(∵$2cosAcosB = cos(A+B)+cos(A-B)$)

$=cos40 +1-(cos60/2) - cos40/2\\=1-cos60/2\\=1-(1/2)/2\\=1-1/4\\=3/4$