Trigonometry?

Question

you have:a+b=90
Prove:
cos(a-b)=2sin(a)*sin(b)

Answer 1

cos(a-b)=2sin(a)*sin(b)
LHS = cos(a)*cos(b) + sin(a)*sin(b)
LHS = cos(90 -b)*cos(90 -a) + sin(a) *sin(b)
LHS = sin(b) * sin(a) + sin(a) *sin(b)
LHS = 2sin(a)*sin(b) = RHS

Answer 2

remember, if a + b = 90, then cos a = sin b, and sin a = cos b. for example: sin 60 = cos 30, sin 45 = cos 45, etc.

now,

cos (a - b) = cos a cos b + sin a sin b

because cos a = sin b and cos b = sin a, we can substitute it into the above expression so that:

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

Answer 3

Hi there!


this answer relies on a couple of facts which you should memorise:
recall that cos(a-b) = cos(a)cos(b) - sin(a)sin(b)
also cos(90-x) = sin(x)

anyway, lets work with the LHS;

LHS = cos(a-b)
= cos(a)cos(b) + sin(a)sin(b)

from the eqn a+b=90, we can see that a=b-90. so continuing on:

= cos(90-b)cos(b) + sin(a)sin(b)

from the eqn a+b=90, we can also see that b=a-90. so continuing on:

= cos(90-b)cos(90-a) + sin(a)sin(b)
= sin(b)sin(a) + sin(a)sin(b)
= 2sin(a)sin(b)
= RHS


hope this helps!!

Answer 4

cos(a-b)=cos(a) cos(b) + sin(a)sin(b) --------(1)

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

subtract (2) from (1)

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

but a+b = 90, so cos(a+b) = cos(90) = 0

so cos(a-b) - 0 = 2 sin(a) sin(b)

cos(a-b) = 2 sin(a) sin(b)

Answer 5

cos(a-b) = cos(a)cos(b) + sin(a)sin(b)
= cos(a)cos(90-a) + sin(a)sin(90-a)
= cos(a)sin(a)+sin(a)cos(a)
= 2cos(a)sin(a)
=2sin(90-b)sin(a)
=2sin(b)sin(a)