# How do you verify the identity sin(pi/2 + x) = cosx?

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

84
Tom Share
Feb 21, 2017

for the "true" proof you need to use matrice, but this is acceptable :

$\sin \left(a + b\right) = \sin \left(a\right) \cos \left(b\right) + \cos \left(a\right) \sin \left(b\right)$

$\sin \left(\frac{\pi}{2} + x\right) = \sin \left(\frac{\pi}{2}\right) \cdot \cos \left(x\right) + \cos \left(\frac{\pi}{2}\right) \cdot \sin \left(x\right)$

$\sin \left(\frac{\pi}{2}\right) = 1$
$\cos \left(\frac{\pi}{2}\right) = 0$

So we have :

$\sin \left(\frac{\pi}{2} + x\right) = \cos \left(x\right)$

Since this answer is very usefull for student here the full demonstration to obtain

$\sin \left(a + b\right) = \sin \left(a\right) \cos \left(b\right) + \cos \left(a\right) \sin \left(b\right)$

(do not read this if you are not fan of math)

a complex numbers can be written in trigonometrics form

$z = \left(\cos \left(x\right) + i \sin \left(x\right)\right)$ $\to \left(1\right)$

multiplying $z$ by $i$ you have

$i z = - \sin \left(x\right) + i \cos \left(x\right)$

because ${i}^{2} = i \cdot i = - 1$

just for you to know, multiplying a complex numbers by $i$ is the same to do a 90° rotation on the complex plane

another way to do a 90° rotation is to derivate $z$

$z ' = - \sin \left(x\right) + i \cos \left(x\right)$

we have

$z ' = i z$

$\frac{z '}{z} = i$

integrating both part

$\ln \left(z\right) = i x + C$

$z = {e}^{i x} {e}^{C}$

taking $x = 0$ and comparing with $\left(1\right)$ you see that C must be $= 0$

so $z = {e}^{i x}$

${e}^{i x} = \cos \left(x\right) + i \sin \left(x\right)$

multiplying by another complex number

${e}^{i x} {e}^{i {x}_{0}} = \left(\cos \left(x\right) + i \sin \left(x\right)\right) \left(\cos \left({x}_{0}\right) + i \sin \left({x}_{0}\right)\right)$

e^(ix)e^(ix_0) = e^(i(x+x_0)

${e}^{i \left(x + {x}_{0}\right)} = \cos \left(x + {x}_{0}\right) + i \sin \left(x + {x}_{0}\right)$

(cos(x+x_0)+isin(x+x_0) = (cos(x)+isin(x))(cos(x_0)+isin(x_0))

develop

(cos(x+x_0)+isin(x+x_0) = cos(x)cos(x_0)+icos(x)sin(x_0) + isin(x)cos(x_0) - sin(x)sin(x_0)

real part of left must be equal to real part of right idem for imaginary part

$\sin \left(x + {x}_{0}\right) = \cos \left(x\right) \sin \left({x}_{0}\right) + \sin \left(x\right) \cos \left({x}_{0}\right)$

note :

$\sin \left(x - {x}_{0}\right) = - \cos \left(x\right) \sin \left({x}_{0}\right) + \sin \left(x\right) \cos \left({x}_{0}\right)$

because $\sin \left(- x\right) = - \sin \left(x\right)$ and $\cos \left(- x\right) = \cos \left(x\right)$

Then teach the underlying concepts
Don't copy without citing sources
preview
?

#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

12
Jun 28, 2017

If you want a simple conceptual way
(rather than a formal way)

#### Explanation:

$\sin \left(\theta\right)$, $\cos \left(\theta\right)$ and 1 are all sides of a right angle triangle.
With $\theta$ being one of the angles that isn't a right angle.

Let's say the angle opposite $\cos \left(\theta\right)$ is called $\phi$

Now, you probably know that there are $\pi$ radians in a triangle (usually stated as 180 degrees in a triangle)

Therefore, $\theta$ + $\phi$ + RightAngle = $\pi$
The right angle is $\frac{\pi}{2}$ so...... $\theta + \phi = \frac{\pi}{2}$
which means,........ $\theta = \frac{\pi}{2} - \phi$

Keep that in the back of your mind

$\cos \left(\theta\right) = \sin \left(\phi\right)$ and $\cos \left(\phi\right) = \sin \left(\theta\right)$
Hopefully you can see why that is?

sub in our formula for $\theta$

$\cos \left(\phi\right) = \sin \left(\frac{\pi}{2} - \phi\right)$
( which also means, $\cos \left(- \phi\right) = \sin \left(\frac{\pi}{2} - \left(- \phi\right)\right)$ )

cos is an even function (meaning $\cos \left(\phi\right) = \cos \left(- \phi\right)$)

so $\cos \left(\phi\right) = \cos \left(- \phi\right) = \sin \left(\frac{\pi}{2} - \left(- \phi\right)\right)$

or put in a better way,...... $\sin \left(\frac{\pi}{2} + \phi\right) = \cos \left(\phi\right)$

I hope this was a little more intuitive than diving into aspects of complex analysis.

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