Expressions for sin(arctan x) and cos(arctan x) Without Trig Functions

If you want to remove the trigonometric functions from $\sin(\arctan x)$ and $\cos(\arctan x)$, the expressions are

$$\sin(\arctan x)=\frac{x}{\sqrt{1+x^2}}, \qquad \cos(\arctan x)=\frac{1}{\sqrt{1+x^2}}.$$

These formulas hold for every real number $x$. The key detail is that $\arctan x$ means the principal inverse tangent, whose values lie in $(-\pi/2,\pi/2)$. That range makes the cosine positive and fixes the sign of the square root.

Derive the formulas with a triangle

Set $\theta=\arctan x$. By the definition of inverse tangent,

$$\tan\theta=x.$$

Think of tangent as the ratio of the opposite side to the adjacent side of a right triangle. Choose the adjacent side to have length $1$ and the opposite side to have length $x$. If $x$ is negative, this is a signed ratio rather than a literal negative side length; the same relationship can be understood using the signs of the trigonometric functions in the angle’s quadrant.

The hypotenuse has length $\sqrt{1+x^2}$, because the Pythagorean theorem gives

$$1^2+x^2=\text{hypotenuse}^2.$$

Therefore, the ratios are

$$\sin\theta=\frac{x}{\sqrt{1+x^2}}, \qquad \cos\theta=\frac{1}{\sqrt{1+x^2}}.$$

Substituting $\theta=\arctan x$ gives the stated expressions. Equivalently, these results follow from $1+\tan^2\theta=\sec^2\theta$, together with the fact that $\cos\theta>0$ on the principal range of arctangent.

Worked example: $x=-3$

Since $\arctan(-3)$ is between $-\pi/2$ and $0$, its sine is negative and its cosine is positive. Substituting $x=-3$ gives

$$\sin(\arctan(-3))=\frac{-3}{\sqrt{1+(-3)^2}}=-\frac{3}{\sqrt{10}},$$

and

$$\cos(\arctan(-3))=\frac{1}{\sqrt{1+(-3)^2}}=\frac{1}{\sqrt{10}}.$$

As a quick check, their ratio is $(-3/\sqrt{10})/(1/\sqrt{10})=-3$, which matches the tangent of the angle. Also, their squares add to $9/10+1/10=1$, as sine and cosine should.

Why the square-root sign matters

From the identity $\sin^2\theta+\cos^2\theta=1$ and $\tan\theta=x$, you can first obtain

$$\cos^2\theta=\frac{1}{1+x^2}.$$

Taking square roots gives $|\cos\theta|=1/\sqrt{1+x^2}$, not automatically $\cos\theta=1/\sqrt{1+x^2}$. The principal angle $\arctan x$ always lies in a range where cosine is positive, so the positive root is correct. Then $\sin\theta=\tan\theta\cos\theta$ gives the sine formula, including its sign.

This sign issue is easy to miss if you start with the Pythagorean identity alone. The range of the inverse function supplies information that the squared identity does not.

Common mistakes

• Using $\sqrt{x^2}$ as $x$. In general, $\sqrt{x^2}=|x|$. Here the denominator is $\sqrt{1+x^2}$, which is positive; the sign of the sine comes from the numerator $x$.
• Giving cosine a negative sign for negative $x$. A negative input makes $\arctan x$ negative, but that angle is still between $-\pi/2$ and $0$, where cosine is positive.
• Confusing $\arctan x$ with $1/\tan x$. The notation means the inverse function of tangent, not the reciprocal. The reciprocal is $\cot x$ when defined.
• Ignoring the principal-value convention. The tangent function repeats, so many angles have tangent $x$. The inverse function $\arctan$ selects the angle in $(-\pi/2,\pi/2)$; using a different angle with the same tangent can change the sine and cosine signs.

For an expression $\arctan(u)$

The same reasoning works whenever $u$ is real. Replace $x$ by $u$:

$$\sin(\arctan u)=\frac{u}{\sqrt{1+u^2}}, \qquad \cos(\arctan u)=\frac{1}{\sqrt{1+u^2}}.$$

For example, if $u=2t-1$, then $\cos(\arctan(2t-1))=1/\sqrt{1+(2t-1)^2}$. No separate sign case is needed for this cosine; for the sine, the sign follows from whether $u$ is positive, zero, or negative.