Proof for arcsin:
The sum and difference of inverse trigonometric functions. Proof for arcsines | MATHVOX
Proof for arccos (Generated by Google Gemini):
To prove the inverse cosine ( \arccos ) sum and difference identities, substitute variables for inverse functions and apply standard trigonometric angle-addition or angle-subtraction formulas. [1, 2]
The general sum and difference formulas for \arccos are:
\arccos (x)\pm \arccos (y)=\arccos \left(xy\mp \sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)
Proof for the Sum Formula: \arccos(x) + \arccos(y)
Step 1: Assign variables to the inverse functions.
Let \alpha = \arccos(x) and \beta = \arccos(y).
By definition, \cos(\alpha) = x and \cos(\beta) = y, where x, y \in [-1, 1] and \alpha, \beta \in [0, \pi].
Step 2: Express \sin(\alpha) and \sin(\beta).
Using the Pythagorean identity \sin^2\theta + \cos^2\theta = 1, we know \sin\theta = \sqrt{1 - \cos^2\theta}.
Because both \alpha and \beta are strictly in the range [0, \pi], the sine is always positive or zero in this domain.
Therefore:
\sin(\alpha) = \sqrt{1-x^2}
\sin(\beta) = \sqrt{1-y^2} [1, 2]
Step 3: Apply the cosine sum formula.
Recall the trigonometric angle-addition identity:
\cos (\alpha +\beta )=\cos (\alpha )\cos (\beta )-\sin (\alpha )\sin (\beta )
Step 4: Substitute the expressions from Step 1 and Step 2.
\cos (\alpha +\beta )=(x)(y)-(\sqrt{1-x^{2}})(\sqrt{1-y^{2}})
Step 5: Take the inverse cosine of both sides.
\alpha +\beta =\arccos \left(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)
Substituting back \alpha = \arccos(x) and \beta = \arccos(y) gives the final identity:
\arccos (x)+\arccos (y)=\arccos \left(xy-\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)
Proof for the Difference Formula: \arccos(x) - \arccos(y)
Step 1: Assign variables.
As before, let \alpha = \arccos(x) and \beta = \arccos(y), meaning \cos(\alpha) = x and \cos(\beta) = y.
Consequently, \sin(\alpha) = \sqrt{1-x^2} and \sin(\beta) = \sqrt{1-y^2}. [1]
Step 2: Apply the cosine difference formula.
Recall the trigonometric angle-subtraction identity:
\cos (\alpha -\beta )=\cos (\alpha )\cos (\beta )+\sin (\alpha )\sin (\beta )
Step 3: Substitute the variables.
\cos (\alpha -\beta )=(x)(y)+(\sqrt{1-x^{2}})(\sqrt{1-y^{2}})
Step 4: Take the inverse cosine of both sides.
\alpha -\beta =\arccos \left(xy+\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)
Substituting back the original definitions yields:
\arccos (x)-\arccos (y)=\arccos \left(xy+\sqrt{1-x^{2}}\sqrt{1-y^{2}}\right)