Expression for $1-\sin x$

Question

Do we take $1-\sin x$ as $(\sin \frac{x}{2}-\cos \frac{x}{2})^2$ or $(\cos \frac{x}{2}-\sin \frac{x}{2})^2$. My teacher says that in most of the cases we take the latter. In one problem the interval was specified as something like $(\frac{\pi}{2},\pi)$ and my teacher told me to take the former. I am facing problem in understanding this. Can someone shed some light on this? Also what about $1+\sin x$.

Both the expressions may be the same but they yield different results (in differentiation and integration problems) and I have myself seen this.

Answer 1

Those expressions are equal so... Both. Regarding the last question, plug $x \mapsto -x$ and you'll have an almost equal expression.

Answer 2

Use that $$\sin^2(x/2)+\cos^2(x/2)-2sin(x/2)\cos(x/2)=1-\sin(x)$$ and $$(\sin(x/2)+\cos(x/2))^2=1+\sin(x)$$

Answer 3

For any real number $x$, you have that $$x^2 = (-x)^2 $$

Since $$\sin\frac x2 - \cos \frac x2 = -\left(\cos \frac x2 - \sin\frac x 2\right)$$

The two expressions, (1): $$\left(\sin\frac x2 - \cos \frac x2\right)^2 $$ and (2): $$\left(-\left(\cos \frac x2 - \sin\frac x 2\right)\right)^2 = \left(\cos\frac x2 - \sin\frac x 2\right)^2 $$ are equal.

Since they are the same, they cannot yield different results. An error must have been made. Unless I'm misunderstanding what exactly is being asked.

Answer 4

Hint:

As for real $a,\sqrt{a^2}=|a|$ which will be $=+a$ if $a\ge0$

Now $1-\sin2y=(\cos y-\sin y)^2=2\sin^2\left(\dfrac\pi4-y\right)$

Now $\sin u\ge0$ if $u$ lies in the first & second quadrant

Similarly, $(\sin y+\cos y)^2=?$