Prove --:--
(i) Sin 20° Sin 40° Sin 60° Sin 80° = 3/16
(ii) Cos 10° Cos 50° Cos 60° Cos 70° = √3/16
Please answer it correctly..........if anybody will write useless and irrelevant thin in the answer he?she will be reported
Answers
Step-by-step explanation:
sin20° sin40° sin60° sin80°
=>. sin 60° sin20° sin40° sin80°
=>. √3/2 sin 20° sin40° sin80°
=>. √3/2 ×1/2 [2sin20° sin40°] sin80°
=>. √3/4 [ cos(40-20) - cos(20+40) ] sin 80°
=>. √3/4 [cos20° - cos60°] sin80°
=>. √3/4 cos20°sin80° - √3/4 cos60°sin80°
=>. √3/4 × 1/2 [2 cos20° sin80°] - √3/4×1/2 sin80°
=>. √3/8 [sin(20+80) -sin(20-80)] - √3/8 sin80°
=>. √3/8 [sin100° - sin(-60°)] - √3/8 sin80°
=>. √3/8 sin100° - √3/8 sin(-60°) - √3/8 sin80°
=>. √3/8 sin(180 - 80) + √3/8 × √3/2 - √3/8 sin80°
=>. √3/8 sin80° + 3/16 - √3/8 sin80°
=>. 3/16 hence proved....
A) By the product-to-sum formula for cosine:
cos(A)cos(B) = (1/2)[cos(A + B) + cos(A - B)].
So, we have:
cos(10°)cos(30°)cos(50°)cos(70°)
= [cos(70°)cos(10°)][cos(50°)cos(30°)]
= (1/2)(1/2)[cos(80°) + cos(60°)][cos(80°) + cos(20°)], from above
= (1/4)[cos^2(80°) + cos(80°)cos(20°) + cos(80°)cos(60°) + cos(60°)cos(20°)].
Using the above formula again:
(1/4)[cos^2(80°) + cos(80°)cos(20°) + cos(80°)cos(60°) + cos(60°)cos(20°)]
= (1/4)(1/2)[cos(160°) + cos(100°) + cos(60°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) + 1]
= (1/8)[cos(160°) + cos(100°) + cos(60°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) + 1]
= (1/8)[cos(160°) + cos(100°) + cos(140°) + cos(20°) + cos(80°) + cos(40°)] + 3/16.
We now want to show that:
cos(160°) + cos(100°) + cos(140°) + cos(20°) + cos(80°) + cos(40°) = 0.
To do this, re-arrange the terms as follows:
[cos(160°) + cos(20°)] + [cos(140°) + cos(40°)] + [cos(100°) + cos(80°)].
Using the sum-to-product formula:
cos(A)cos(B) = 2cos[(A + B)/2]cos[(A - B)/2],
each bracketed term equals zero (as (A + B)/2 = 90° and cos(90°) = 0), so this equals:
0 + 0 + 0 = 0, as required.
Therefore, cos(10°)cos(30°)cos(50°)cos(70°) = (1/8)(0) + 3/16 = 3/16.