Dewie B.

asked • 03/17/20# Solve equation for unknown angle

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.

## 3 Answers By Expert Tutors

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)√(a^{2} + b^{2}). 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.

Rich G. answered • 03/17/20

Experienced Pre-Calculus/Trigonometry Tutor

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

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

Dewie B.

That's a good try at it. I never would have thought to square both sides. But I tried plugging your answer into the original equation and it didn't work.03/17/20

Carmine F.

03/17/20

Carmine F.

03/17/20

## Still looking for help? Get the right answer, fast.

Get a free answer to a quick problem.

Most questions answered within 4 hours.

#### OR

Choose an expert and meet online. No packages or subscriptions, pay only for the time you need.

Dewie B.

I have to give all answers between 0 and 2pi and answers must be exact, if that makes any difference.03/17/20