Choose principal values and valid signs

Choose the principal inverse-trigonometric value

Write the principal angle interval before evaluating. An inverse function gives one selected angle; solving a trigonometric equation may require more angles in the requested interval.

Inverse trigonometric functions recover one selected angle from a ratio. They are not reciprocals of the trigonometric functions. Although infinitely many angles can have the same ratio, the calculator gives the principal value, as the source remark emphasises.

$\sin^{-1}x$

Allowed x: $[-1,1]$

Principal angle interval: $[-\pi/2,\pi/2]$

Printed example and reason: $\sin^{-1}(1/2)=\pi/6$, since $\sin(\pi/6)=1/2$

$\cos^{-1}x$

Allowed x: $[-1,1]$

Principal angle interval: $[0,\pi]$

Printed example and reason: $\cos^{-1}(1/\sqrt2)=\pi/4$, since $\cos(\pi/4)=1/\sqrt2$

$\tan^{-1}x$

Allowed x: All real x

Principal angle interval: $(-\pi/2,\pi/2)$

Printed example and reason: $\tan^{-1}\sqrt3=\pi/3$, since $\tan(\pi/3)=\sqrt3$

The three source inverse graphs retain their endpoints and monotonic directions. Inverse tangent approaches horizontal asymptotes at ±π/2 and never reaches them.

Examples 17–18 apply these principal ranges before evaluating an outer function or choosing a triangle’s height. Continue with Tutorial Question 13.

Keep the identity denominators and square-root signs

Choose an identity that contains the expression you need. Before dividing, exclude zeros of the divisor; before taking a square root, use the angle interval to choose the sign.

The basic identity is $\sin^2\theta+\cos^2\theta=1$. Dividing by cos²θ, where cosθ≠0, gives $\tan^2\theta+1=\sec^2\theta$. Dividing by sin²θ, where sinθ≠0, gives $1+\cot^2\theta=\operatorname{cosec}^2\theta$. Example 19 retains the acute degree interval that lets $\sqrt{\sec^2\theta}=\sec\theta$; generally the square root is $|\sec\theta|$.