# Expressing $\sin(\alpha)$ in terms of $\cos$, and $\cos \alpha$ in terms of $\sin$

When we express $\sin$ to $\cos$ or from $\cos$ to $\sin$:

why do we want to subtract the angle ($\alpha$) from ($\pi/2$)???

• i know that the angle btw sin and cos is (pie/2) ,but why we need to subtract ??!!!! – Zahraa Haj Hassan Sep 18 '16 at 14:52
• are you asking why $\cos x=\sin(\frac \pi2-x)$? – Hagen von Eitzen Sep 18 '16 at 14:52
• There are no such hard and fast rules. It depends upon the identity or theorem one is trying to prove, or the expression one is trying to simplify. Knowing when to change $sin$ into $cos$ or vice versa, only comes from intuition and experience which can be built by solving a large variety of problems. Solving a lot of problems will make your brain better at recognizing patterns and you will thus develop a "feel" for when to, what to and why to transform. Slowly you will start realizing the usefulness of transformations without actually performing them – Anonymous_original Sep 18 '16 at 14:59
• Mmmm... Pie.... – Deepak Sep 18 '16 at 15:04
• "cosine" is short for "complementary sine". The cosine of an (acute) angle is, by definition, "the sine of the complementary angle". – Blue Sep 18 '16 at 16:14

Because $\sin(x)=\cos(\pi/2-x)$.

I really don't know what to say beyond this.

Perhaps looking at the angles in a right triangle might give you some insight.

• I need the reason the main reason😖 – Zahraa Haj Hassan Sep 18 '16 at 15:24
• @ZahraaHajHassan Are you asking why $\cos x=\sin(\frac \pi2-x)$??? – Anonymous_original Sep 18 '16 at 15:28

I don't know whether this answers the question, but I've decided to have a go at it anyway.

Let us consider the $rt. \angle d$ $\Delta$ in the figure above.
Here by definition, $$\sin\Theta = \frac{o}{h}$$ $$\cos\Theta = \frac{a}{h}$$

Now, clearly the $3^{rd} \angle$ of the $\Delta$ is $\frac{\pi}{2} - \Theta$.

If we now take into consideration the $T - Ratios$ of $\frac{\pi}{2} - \Theta$, we get, $$\sin(\frac{\pi}{2} - \Theta) = \frac{a}{h}$$ $$\cos(\frac{\pi}{2} - \Theta) = \frac{o}{h}$$

$\therefore$ by comparing the values of $\sin\Theta$, $\cos\theta$, $\sin(\frac{\pi}{2} - \Theta)$ and $\cos(\frac{\pi}{2} - \Theta)$, we get, $$\sin(\frac{\pi}{2} - \Theta) = \cos\Theta$$ $$\cos(\frac{\pi}{2} - \Theta) = \sin\Theta$$

• Any way thank you very much this a good proof👍 – Zahraa Haj Hassan Sep 18 '16 at 16:39