How do I prove: $\cos (\theta + 90^\circ) \equiv - \sin \theta $ [duplicate]

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How do I go about proving this? I know one method is:

$\eqalign{ \cos (90^\circ + \theta ) &\equiv \cos90^\circ \cos\theta - \sin90^\circ \sin\theta \cr & \equiv (0)(\cos\theta ) - (1)(\sin \theta ) \cr & \equiv - \sin \theta \cr} $

I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.

Thanks alot.

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6 Answers

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I'd appreciate others, particularly ones that will allow me to visualize this identity on the unit circle.

The picture represents the trigonometric unit circle. The angle $\theta=\angle AOP$ is represented in the first quadrant, but it is a generic orientate angle. $|AP|=\sin\theta$. $\angle AOQ=\theta+90^\circ$. The right triangle $[O,A,P]$ is similar to the right triangle $[O,B,Q]$, because $\angle BQO=\angle AOP=\theta$. Since $|OP|=|OQ|=1$, both triangles are congruent and $$\cos(\theta+90^\circ)=-|OB|=-|AP|=-\sin\theta.$$

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The point $P$ on the unit circle corresponding to $\theta$ is $\langle\cos\theta,\sin\theta\rangle$. Assume for a moment that $P$ is not on either coordinate axis; then it’s on the line $y=(\tan\theta)x=\frac{\sin\theta}{\cos\theta}x$.

The point $Q$ corresponding to $\theta+\frac{\pi}2$ is on the perpendicular line through the origin, whose slope is $-\frac{\cos\theta}{\sin\theta}$. But $Q$ is also the point $\left\langle\cos\left(\theta+\frac{\pi}2\right),\sin\left(\theta+\frac{\pi}2\right)\right\rangle$, so

$$\frac{\sin\left(\theta+\frac{\pi}2\right)}{\cos\left(\theta+\frac{\pi}2\right)}=-\frac{\cos\theta}{\sin\theta}\;.\tag{1}$$

The two sides of $(1)$ are evidently the two square roots of $\cot^2\theta$, and they have the additional property that the squares of numerator and denominator sum to $1$, so they must be identical except for sign. Thus, either

$$\cos\left(\theta+\frac{\pi}2\right)=\sin\theta\quad\text{and}\quad\sin\left(\theta+\frac{\pi}2\right)=-\cos\theta\;,$$

or

$$\cos\left(\theta+\frac{\pi}2\right)=-\sin\theta\quad\text{and}\quad\sin\left(\theta+\frac{\pi}2\right)=\cos\theta\;,$$

and consideration of quadrants rules out the former. The case in which $P$ is on a coordinate axis is easily handled separately.

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Remember that sine and cosine are co-functions, meaning that they are connected through complementary angles (this is evident for $\newcommand{\dg}{^\circ} 0\dg \le \theta \le 90\dg$ from geometry of a right triangle). $$\begin{align} \cos \theta &= \sin(90\dg - \theta)\\ \sin \theta &= \cos(90\dg - \theta) \end{align} $$

Now, sine is an odd function, which means that $$\sin(-\theta) = -\sin\theta \qquad \text{for all } \theta.$$

Putting these together:

$$ \cos(\theta + 90\dg) = \cos(90\dg - (-\theta)) = \sin(-\theta) = -\sin\theta. $$

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Using Euler's formula we get: $$\cos(90 + \theta) = \cos (\pi/2 + \theta) = \frac{e^{i(\pi/2 + \theta)}+e^{-i(\pi/2 + \theta)}}{2} = \frac{e^{i\theta} e^{i\pi/2}+e^{-i\theta} e^{-i\pi/2}}{2} = \frac{ie^{i\theta} - ie^{-i\theta}}{2} = -\frac{e^{i\theta} - e^{-i\theta}}{2i} = -\sin(\theta) $$

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$\sin$ is the vertical position. $\cos$ is the horizontal position. If you add 90 to the angle $\theta$, that's equivalent to rotating the coordinate system by 90°: the axes switch places.

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So you have cos(x+90)=-sin(x) You know that cos(90-x)=sin(x), and hopefully this make sense why (if not, imagine a right triangle with angles x and 90-x. Take the sin x and cos(90-x) and you will see they will be the same). You also know that cos (w)=cos(-w) (you can see why this is by imagining w and -w on the unit circle). So let's factor a -1 from cos(90-x). We are left with cos(x-90)=-sin(x). Now remember that cos(x)=-cos(x+180). So let's apply this to our problem. -cos(x-90+180)=sin(x), so cos(x+90)=-sin(x)

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