sinA=.....+cosA
I did this yesterday
http://www.jiskha.com/display.cgi?id=1459511832
Ok, how about this approach
sinA = x + cosA
x = sinA - cosA
let's change sinA - cosA into a single sinusoidal expression:
Let sinA - cosA = a sin(A + k)
a sin(A+k)
= a(sinAcosk + cosAsink_
= asinAcosk + acosAsink
then asinAcosk = sinA
acosk = 1
cosk = 1/a
acosAsink = -cosA
asink = -1
sink = -1/a
but we know sin^2 k + cos^2 k = 1
(-1/a)^2 + (1/a)^2 = 1
2/a^2 = 1
a^2 = 2
a = sqrt(2)
also tank = sink/cosk = (-1/a) / (1/a) = -1
k = -Pi/4
so sinA - cosA = sqrt(2) sin(A - Pi/4)
x = sqrt(2) sin(A - Pi/4)
so I plotted on Wolfram:
y = sinx - cosx and y = sqrt(2) sin(x - Pi/4)
and the two curves coincides
https://www.wolframalpha.com/input/?i=plot+y+%3D+sinx+-+cosx,+y+%3D+sqrt(2)+sin(x+-+Pi%2F4)
To find the value of sinA in terms of cosA, you can use the Pythagorean identity:
sin^2(A) + cos^2(A) = 1
Rearranging this equation, we get:
sin^2(A) = 1 - cos^2(A)
Taking the square root of both sides, we get:
sin(A) = √(1 - cos^2(A))
So, sinA is equal to the square root of 1 minus cos^2(A).