Expression of sine

Question

Can someone help me explain why it is true that

$$\sin(\pi/2-\theta)=\sqrt{1-\sin^2\theta}$$

When answering please explain the different relation which is used

Thanks

Answer 1

You should be asking why

$$\sin^2(\pi/2-\theta) = 1- \sin^2(\theta).$$

By the trigonometric Pythagorean theorem, we know that

$$\cos^2(\theta)+\sin^2(\theta)=1 \implies \cos^2(\theta)=1-\sin^2(\theta)$$

is valid.

If you additionally use the complementary formula for trigonometric functions $\cos(\theta)=\sin(\pi/2-\theta)$ then you can conclude what you wanted to show in the first place.

The reason why $$\sin(\pi/2-\theta) = \sqrt{1- \sin^2(\theta)}$$

is not always true is because you dropped the minus sign. The sign must be chosen according to the quadrant in which the angle is located.

EDIT: If $-\pi/2\leq \theta \leq \pi/2$ (first and fourth quadrant) then the sign is positive because the cosine function is positive for this interval. If $\pi/2 < \theta <3\pi/2$ (second and third quadrant) then the sign is negative because the cosine function is negative for this interval.

Answer 2

Simply note that

• $1-\sin^2\theta=\cos^2 \theta$
• $\sin \left(\frac{\pi}2-\theta\right)=\cos \theta$

thus, since RHS is non negative

$$\sin \left(\frac{\pi}2-\theta\right)=\sqrt{1-\sin^2\theta}$$

is true if and only if $\sin \left(\frac{\pi}2-\theta\right)=\cos \theta\ge0$ that is $\theta \in\left[-\frac{\pi}2+2k\pi,\frac{\pi}2+2k\pi\right]$.

Answer 3

$$ \sin \left(\frac{\pi}2-\theta\right) =\cos\theta \not = |\cos\theta| =\sqrt{\cos^2\theta}=\sqrt{1-\sin^2\theta}$$