Solve equation for unknown angle

Question

I've been solving equations like 2(sinx)^2+sinx=1 that can be done using factoring and that's not too hard. But now I have an equation sinx-cosx=1/2 and nothing is squared so I don't know how to even start. Help please.

Answer 1

For linear combinations of sine and cosine, we have the identity

a sin(x) + b cos (x) = c sin (x + tan^-1(b/a))

where c = sgn(a)√(a² + b²). In your case, a = 1 and b = -1, so that c = √2 and tan^-1(b/a) = -π/4. This allows us to rewrite your equation as

√(2)sin(x - π/4) = 1/2

and hence

sin(x - π/4) = √(2)/4.

This gives the exact solutions

x = π/4 + sin^-1((√2)/4) + 2πn

as well as

x = 5π/4 - sin^-1((√2)/4)+ 2πn

where n is any integer. The only solutions that lie between 0 and 2π occur when n = 0.

Answer 2

This is a tough one, it took me a few tries to find the right identity to solve it. Solving it graphically was easy though.

If we divide both sides by √2 we get

¹/_√2 sin x - ¹/_√2 cos x = ¹/_2√2

Since ¹/_√2 = cos ^π/₄ = sin ^π/₄ we can rewrite this equation as

cos ^π/₄ sin x - sin ^π/₄ cos x = ¹/_2√2

Which we can reorganize as

sin x cos ^π/₄ - cos x sin ^π/₄ = ¹/_2√2

From the angle-difference identity sin (α-β) = sin α cos β - cos α sin β, we can rewrite the above as

sin (x - ^π/₄) = ¹/_2√2

Take the inverse side of both sides and you get

(x - ^π/₄) = sin^{-1 1}/_2√2

(x - ^π/₄) ≈ .3614

x ≈ 1.1467

Answer 3

Let st = sin(theta) and ct = cos(theta) then

st-ct=1/2

(st-ct)^2=(1/2)^2 , by squaring both side

st^2+ct^2-2stct=1/4 , by FOIL

1-2stct=1/4 , by Pythagorean theorem

2stct=3/4 , by collecting like terms

s(2t)=3/4 , by double-angle formula

2t=arcsin(3/4)

t=arcsine(3/4)/2

~appx.= 48.59/2 = 24.295

hope this helps!

best,

Carmine