sin(x) + cos(x) = sin(x) + sqrt(1 – sin^2(x))

Well the first term is already sin(x), so we just need to express cos(x) in terms of sin(x).

You know the identity sin^2(x) + cos^2(x) = 1, right? Well solving this for cos(x) and substituting it into the original expression gives

sin(x) + ±√[1 – sin^2(x)], which is

sin(x) ± √[1 – sin^2(x)]

(sin(x) + cos(x))^2= sin(x)^2+cos(x)^2+2sin(x)cos(x)

= 1+sin(2x)

Remember that cos(x)^2 + sin(x)^2 = 1, and solve for cos(x).

cos(x)^2 + sin(x)^2 = 1

cos(x)^2 = 1 – sin(x)^2

cos(x) = sqrt(1 – sin(x)^2)Now just plug that in for cos(x) in the original expression to get sin(x) + sqrt(1 – sin(x)^2).

I am sure you would like it written in terms of sine only without nasty square roots.

asinx + bcosx

sinx + cosx

a = 1, b = 1ksin(x – α)

k(sinxcosα – cosxsinα)

ksinxcosα – kcosxsinα

kcosαsinx – ksinαcosx

sinx + cosx

kcosα = 1, ksinα = 1r = √(a² + b²)

r = √(1² + 1²)

r = √2tana = ksina / kcosa

tana = 1

Sin is positive and cos is positive so a is in quadrant 1

a = atan1

a = 45°sinx + cos x = √2sin(x + 45)

sin² x +cos² x=1

cos² x=1-sin² x

cosx=(√(1-sin² x))

so I getsinx+(√(1-sin² x))