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In proving $\sin(\pi/2 - x) = \cos(x)$, in my book its given that

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

So I understand this but im confused about how $\cos x$ is being obtained from $\cos(\pi/2-(\pi/2-x))$. Please explain it to me.

PrincessEev
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Mad Dawg
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    I made a quick edit to the title of your question. Based on the others' answers, it seems there was confusion as to what you were looking for. Figured it best to let you know about this edit just in case you have a problem with it or anything of the sort. :) – PrincessEev Apr 06 '19 at 10:43
  • Yeah I was thinking to edit it to, thanks – Mad Dawg Apr 06 '19 at 10:44

4 Answers4

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If you distribute the minus sign you see that

$$\frac \pi 2 - \left( \frac \pi 2 - x \right) = \frac \pi 2 - \frac \pi 2 + x = x$$

PrincessEev
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\begin{align*} \sin\left(\frac{\pi}{2}-x\right) &= \left(\sin \frac{\pi}{2}\right)\cos x - \cos \frac{\pi}{2} \sin x \\ &= \cos x &\left(\cos \frac{\pi}{2} = 0 \text{ and } \sin \frac{\pi}{2} =1 \right) \end{align*}

Theo Bendit
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Vikram
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$$\cos \left(\dfrac{\pi}{2} - \left(\dfrac{\pi}{2} - x \right) \right) = \cos \left(\dfrac{\pi}{2} - \dfrac{\pi}{2} + x \right) = \cos(x) $$

PrincessEev
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Finn Eggers
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The most simple answer will be the phase difference between $\sin x$ and $\cos x$ is $\frac{\pi}{2}$. The point is how trigonometric functions are defined ?

The fact is they are never defined rather interpreted in a logical,pedagogical manner. See if you put $x=\frac{\pi}{2}$ and $y=-x$ in $\sin(x+y)=\sin x \cos y+\cos x \sin y$ then you will get the verification.

But how this formula is derived ? Simple euclidean geometry. The only defination you need is what is the meaning of sine,cosine in terms of ratio.

Hey wait ! I have used the particular values of sine and cosine. Though some values can be imagined. I will recommend to check out following links.

How would a triangle for sin 90 degree look

How were the sine, cosine and tangent tables originally calculated?

NewBornMATH
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    This is not what the OP wants. He simply wants to know why $\cos(\pi/2 - (\pi/2 - x)) = \cos(x)$. Directly, not by doing a different method of proof. – PrincessEev Apr 06 '19 at 10:07
  • Oh ! But that's only the distribution property of negative sign. The title of question must be edited. – NewBornMATH Apr 06 '19 at 10:37
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    That's a fair point. I went ahead and made the edit to clarify and made a comment on the question proper just in case. – PrincessEev Apr 06 '19 at 10:44