Question

(1+cosx) (1+sinx) = 5/4 Find a)(1-cosx) and b)(1-sinx)

Answer 1

Let (1 + sinx)(1 + cosx) = K, where K = 5/4
Let (1 - sinx)(1 - cosx) = X.

Multiplying the two equations together,
(1 + sinx)(1 – sinx)(1 + cosx)(1 – cosx) = KX
(1 – sin^2x)(1 – cos^2x) = KX
cos^2x.sin^2x = KX.......(1)
cosx.sinx = √(KX)

Expanding the two equations,
1 + sinx + cosx + sinx.cosx = K
1 – sinx – cosx + sinx.cosx = X
Adding the two equations together
2 + 2.sinx.cosx = K + X
Substituting for cosx.sinx = √(KX)
2 + 2√(KX) = K + X

Setting K = 5/4,
2 + √(5X) = 5/4 + X
8 + 4√(5X) = 5 + 4X
4√(5X) = 4X – 3
16(5X) = 16X^2 – 24X + 9
16X^2 – 24X – 80X + 9 = 0
16X^2 – 104X + 9 = 0

Using the quadratic formula,
X = (104 ± √(104^2 – 4.16.9))/(2.16) = (104 ± √(10,816 – 576))/(32) = (104 ± √(10,240))/32
X = 13/4 ±√(10)
X = 0.087722, or X = 6.412278

But sinx.cosx = √(KX),
and |sinx| < 1 and |cosx| < 1 => KX < 1.

Therefore, X cannot equal 6.412278, so ignore this solution.

Now consider equation (1)

Let sin^2x =y
y(1-y) =KX=(5/4)×(0.08772)=0.109

y-y^2=0.109
y^2-y+0.109=0

y=(1(+-)√(1-4×1×0.109))/2
((1+...) cannot be the solution as y is sin^2)
=(1-√(0.56))/2
=sin^2x=(1-(√0.56))/2
= 0.125
sinx= 0.35
(1-sinx)=1-0.35=0.65

Cosx=√(1-sin^2x)=√(1-0.125)=√0.875=0.935

(1-cosx )= 1-0.935=0.065