Apair of sneakers is on sale for $51. this is 75% of the original price. what was the original price of the shoes?

Hello from MrBillDoesMath!

See Discussion below

Discussion:

(sinq + cosq)^2 =                                     => (a +b)^2 = a^2 + 2ab + b^2

(sinq)^2 + (cosq)^2 + 2 sinq* cosq        => as (sinx)^2 + (cosx)^2 = 1

1 + 2 sinq*cosq              (*)

Setting a = b = q in the trig identity:

sin(a+b) = sina*cosb + cosa*sinb                      

sin(2q) =                      (**)

sinq*cosq + cosq*sinq      => as both terms are identical

2 sinq*cosq

Combining (*) and (**)

(sinq + cosq)^2 = 1 + 2sinq*cosq     => (**) 2sinq*cosq =  sqin(2q)

                          = 1 +  sin(2q)

Hence

(sinq + cosq)^2  = 1 +  sin(2q)            => subtracting 1 from both sides

(sinq + cosq)^2  - 1 =   sin(2q)  

The last statement is what we are trying to prove.

Thank you,

MrB