"sin⁡2θ+cos⁡2θ=sin⁡θ+cos⁡θ proofI tried to do like thisL.H.S.=2sin⁡θcos⁡θ+cos2θ−sin2θ=sin⁡θcos⁡θ+cos2θ+sin⁡θcos⁡θ−sin2θ=cos⁡θ(sin⁡θ+cos⁡θ)+sin⁡θ(cos⁡θ−sin⁡θ)Then what..." was answered by our community

ab0utfallingm1z2

Answered question

2022-03-30

\sin 2 θ + \cos 2 θ = \sin θ + \cos θ

proof
I tried to do like this
L.H.S.

= 2 \sin θ \cos θ + \cos^{2} θ - \sin^{2} θ

= \sin θ \cos θ + \cos^{2} θ + \sin θ \cos θ - \sin^{2} θ

= \cos θ (\sin θ + \cos θ) + \sin θ (\cos θ - \sin θ)

Then what should I do ?

Yaritza Phillips

Beginner2022-03-31Added 12 answers

Let be the equation to solve.

\Rightarrow \sqrt{2} \cos (2 θ - \frac{π}{4}) = \sqrt{2} \cos (θ - \frac{π}{4})

\Rightarrow 2 θ - \frac{π}{4} \equiv θ - \frac{π}{4} (mod 2 π) or 2 θ - \frac{π}{4} \equiv - θ + \frac{π}{4} (mod 2 π)

\Leftrightarrow θ \equiv 0 (mod 2 π) or θ \equiv \frac{π}{6} (mod \frac{2 π}{3})

armejantm925

Beginner2022-04-01Added 20 answers

For

θ = π

you have
LHS:

\sin (2 π) + \cos (2 π) = 0 + 1 = 1

RHS:

\sin π + \cos π = 0 + (- 1) = - 1

hence the equality does not hold and can't be proven.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?