Preccalc inverse functions formulas (like with triangles)

sin(arccosx)

squareroot of(1-x^2)

sin(arcsin x)

x

cos(arcsin x)

sqrt{1-x^{2}}

tan(arcsin x)

frac{x}{sqrt{1-x^{2}}}

sin(arccos x)

sqrt{1-x^{2}}

cos(arccos x)

x

tan(arccos x)

frac{sqrt{1-x^{2}}}{x}

sin(arctan x)

frac{x}{sqrt{x^{2}+1}}

cos(arctan x)

frac{1}{sqrt{x^{2}+1}}

tan(arctan x)

x

How do you derive a composite inverse trig function like cos(arcsin x)?

Substitute: Let θ equal the inner inverse trig function (e.g., θ =arcsin x). Rewrite: The problem becomes the outer trig function of θ (e.g., cos(θ)). Draw a Triangle: Sketch a right triangle and label one acute angle θ. Label Sides: Use the substituted equation (sin θ =x) and the definition of the trig function to label two sides of the triangle. Pythagorean Theorem: Use a^{2}+b^{2}=c^{2} to find the length of the third side. Solve: Use the triangle to find the value of the final expression.

How do you derive the formula for cos(arcsin x)?

Let θ =arcsin x, so sin θ =x=frac{x}{1}. Draw a right triangle with opposite side x and hypotenuse 1. Find the adjacent side using the Pythagorean theorem: adjacent=sqrt{1^{2}-x^{2}}=sqrt{1-x^{2}}. Find the cosine of θ: cos θ =frac{adjacent}{hypotenuse}=frac{sqrt{1-x^{2}}}{1}=sqrt{1-x^{2}}.

How do you derive the formula for sin(arccos x)?

Let θ =arccos x, so cos θ =x=frac{x}{1}. Draw a right triangle with adjacent side x and hypotenuse 1. Find the opposite side: opposite=sqrt{1^{2}-x^{2}}=sqrt{1-x^{2}}. Find the sine of θ: sin θ =frac{opposite}{hypotenuse}=frac{sqrt{1-x^{2}}}{1}=sqrt{1-x^{2}}.

How do you derive the formula for tan(arcsin x)?

Let θ =arcsin x, so sin θ =x=frac{x}{1}. Draw a right triangle with opposite side x and hypotenuse 1. Find the adjacent side: adjacent=sqrt{1-x^{2}}. Find the tangent of θ: tan θ =frac{opposite}{adjacent}=frac{x}{sqrt{1-x^{2}}}.

How do you derive the formula for sin(arctan x)?

Let θ =arctan x, so tan θ =x=frac{x}{1}. Draw a right triangle with opposite side x and adjacent side 1. Find the hypotenuse: hypotenuse=sqrt{x^{2}+1^{2}}=sqrt{x^{2}+1}. Find the sine of θ: sin θ =frac{opposite}{hypotenuse}=frac{x}{sqrt{x^{2}+1}}.

How do you derive the formula for cos(arctan x)?

Let θ =arctan x, so tan θ =x=frac{x}{1}. Draw a right triangle with opposite side x and adjacent side 1. Find the hypotenuse: hypotenuse=sqrt{x^{2}+1}. Find the cosine of θ: cos θ =frac{adjacent}{hypotenuse}=frac{1}{sqrt{x^{2}+1}}.