Prove that (COs x - cosy)² + (sin x - Siby)?
- a S14(x+y)
Answers
Show that:
- ( Cos x + Cos y )² + ( Sin x + Sin y)² = 4cos² (x-y/2)
Step-by-step explanation:
W e know that,
2SinA SinB = Cos(A-B)-Cos (A+B) …(1)
And,
2Cos ACosB = Cos(A+B)+Cos (A-B) …(2)
Using (1) & (2),
We get,
( Cos x + Cos y )² + ( Sin x + Sin y)²
= cosx² + sinx² + 2cosxcosy + siny² + cosy² + 2cosxcosy
=1+1+2cosxcosy+2sinxsiny
[Since,sinA²+cosA²=1]
=2+cos( x+y) +cos(x-y)+cos(x-y)-cos(x+y)
=2+2cos(x-y)
=2+2[{2cos (x-y)²–1}] [N.B-cos2A=2(cosA²)–1]
=2–2+{4cos(x-y/2)²}
={4cos(x-y/2)²}
Proved!!
Answer:
Step-by-step explanation:A Traveler starts from A towards B at 7:00 a.m. and another Traveler starts at the same time from B towards A
Let say Distance between A & B = D km
They meet at 9:00 a.m. at 36 kilometer from A
=> Distance Travelled by A = 36 km
& Distance Travelled by B = D - 36 km
Time taken = 2 hrs
Speed of A = 36/2 = 18 km/hr
Speed of B = (D-36)/2 km/Hr
further Time taken by B to reach A
= 36/ ((D-36)/2)
= 72/(D - 36) hr
Distance covered by A in this time
= 18 * 72/(D - 36)
18 * 72/(D - 36) = D - 36 - 36 ( 36 km Already covered & 36 km still to reach B)
=> 18*72 = (D-72)((D - 36)
=> D² -108D + 36*72 - 18*72 = 0
=> D² - 108D + 18*72 = 0
=> D = (108 ± √((108)² - 4*18*72))/2
=> D = 94.25 km ignoring value < 36 km
Distance between A & B = 94.25 km