In pre-calc, we're doing trigonometric identities, and I have a question but my teacher isn't available right now. We have to verify trigonometric identities, but I can't progress right now because I'm unsure if cotangent squared equals cosine squared over sine squared. I'm not asking for the answer to my problem, I just don't know if my thinking pattern is correct here.
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2$\begingroup$ There is a certainly level in mathematics where it is OK to write $\cot$ with no angle. But you're not at that level, and it's unwise to get into the bad habit of writing $\cot^2$ instead of $\cot^2 x$ or $\cot^2 \theta.$. $\endgroup$– B. GoddardFeb 12, 2019 at 18:16
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4 Answers
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$$\cot^2x=(\cot x)^2=\left(\dfrac{\cos x}{\sin x}\right)^2=\dfrac{\cos^2x}{\sin^2x}$$
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Yes, by definition $\cot x = \frac{1}{\tan x}$, and $\tan x = \frac{\sin x}{\cos x}$, so $$\cot x = \frac{1}{\sin x/\cos x} = \frac{\cos x}{\sin x}.$$ Thus, squaring both sides of the equation, we obtain $\cot^2 x = \frac{\cos^2 x}{\sin^2 x}$.
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3$\begingroup$ It's more usual to define $\cot x = \frac{\cos x}{\sin x}$ so that for example $\cot(\pi/2) = 0$ even though $\tan(\pi/2)$ is undefined. $\endgroup$ Feb 11, 2019 at 22:19
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$\begingroup$ I see, very good. I guess the 'definition' I provide of $\cot x$ is more a case of how I remember it, but the way you suggest is clearly the more appropriate definition. Thanks $\endgroup$ Feb 11, 2019 at 22:41
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Since the cotangent function is defined as cosine divided by sine $($or as reciprocal tangent, which is defined by sine divided by cosine, respectively$)$ we get that
$$\cot^2(x)=(\cot(x))^2=\left(\frac{\cos(x)}{\sin(x)}\right)^2=\frac{(\cos(x))^2}{(\sin(x))^2}=\frac{\cos^2(x)}{\sin^2(x)}$$
$$\therefore~\cot^2(x)~=~\frac{\cos^2(x)}{\sin^2(x)}$$
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$$ \frac{\sin(x)}{\cos(x)} = \tan(x) $$
Thus $$ \frac{\cos(x)}{\sin(x)} = \frac{1}{\tan(x)} = \cot(x) $$ and $$ \Big(\frac{\cos(x)}{\sin(x)}\Big)^{2} = \frac{\cos^{2}(x)}{\sin^{2}(x)} = \cot^{2}(x) $$
