Get the solution to "How do I prove:cos⁡(θ+90∘)≡−sin⁡θ"

Asiya Holder

Answered question

2022-02-28

How do I prove:

\cos (θ + 90^{\circ}) \equiv - \sin θ

aksemaktjya

Beginner2022-03-01Added 6 answers

Remember that sine and cosine are co-functions, meaning that they are connected through complementary angles (this is evident for

0^{\circ} \leq θ \leq 90^{\circ}

from geometry of a right triangle).

\cos θ = \sin (90^{\circ} - θ)

\sin θ = \cos (90^{\circ} - θ)

Now, sine is an odd function, which means that

\sin (- θ) = - \sin θ

for all

θ

Putting these together:

\cos (θ + 90^{\circ}) = \cos (90^{\circ} - (- θ)) = \sin (- θ) = - \sin θ

Tye Rhodes

Beginner2022-03-02Added 4 answers

Using Euler's formula we get:

\cos (90 + θ) = \cos (\frac{π}{2} + θ)

= \frac{e^{i (\frac{π}{2} + θ)} + e^{- i (\frac{π}{2} + θ)}}{2}

= \frac{e^{i θ} e^{i \frac{π}{2}} + e^{- i θ} e^{- \frac{π}{2}}}{2}

= \frac{i e^{i θ} - i e^{- i θ}}{2}

= - \frac{e^{i θ} - e^{- i θ}}{2 i}

= - \sin (θ)

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