How to simplify sin⁡(arcsin⁡x+arccos⁡x)?

Question

Anabel Phillips · Accepted Answer

We can approach this issue in two different ways: first, by using right triangle trigonometry (applicable for positive

x

) and purely trigonometric (applicable for all

x

, but we will use it for

x \leq 0

.
Let's analyze this problem from the position of trigonometry of the right triangle. For this, let's assume that

0 < x \leq 1

.
Then

∠ A = \arcsin x

is an angle from

0

to

\frac{π}{2}

, sine of which is

x

.
Assume, this angle

∠ A

is an acute angle of a right triangle

Δ A B C

. So,

\sin (∠ A)

is a ratio of an opposite cathetus

a

to hypotenuse

c

:

x = \sin (∠ A) = \frac{a}{c}

Consider another acute angle of this triangle -

∠ B

.
Expression

\frac{a}{c}

represents its cosine.
Hence,

∠ B = \arccos (\frac{a}{c}) = \arccos x

Now we see that

∠ A = \arcsin x

and

∠ B = \arccos e x

are two acute angles in a right triangle. Their sum is always

\frac{π}{2}

.
Hence,

\sin (\arcsin x + \arccos x) = \sin (\frac{π}{2}) = 1

Case with non-positive

x

we will consider differently.
If

x \leq 0

,

\arcsin x

is between

- \frac{π}{2}

and

0

.
If

x \leq 0

,

\arccos x

is between

\frac{π}{2}

and

π

.
Using formula for

\sin

of a sum of two angles,

\sin (\arcsin x + \arccos x) =

= \sin (\arcsin x) \cdot \cos (\arccos x) + \cos (\arcsin x) \cdot \sin (\arccos y)

By definition of inverse trigonometric functions

\arcsin

and

\arccos

, we write the following:

\sin (\arcsin x) = x

\cos (\arccos x) = x

Using trigonometric identity

\sin^{2} (ϕ) + \cos^{2} (ϕ) = 1

, we can find the other components:

\cos^{2} (\arcsin x) = 1 - \sin^{2} (\arcsin x) = 1 - x^{2}

\sin^{2} (\arccos x) = 1 - \cos^{2} (\arccos x) = 1 - x^{2}

Since

\arcsin x

is between

- \frac{π}{2}

and

0

, its cosin is positive:

\cos^{2} (\arcsin x) = 1 - x^{2}

\Rightarrow \cos (\arcsin x) \sqrt{1 - x^{2}}

Since

\arccos x

is between

\frac{π}{2}

and

π

, its sine is positive:

\sin^{2} (\arccos x) = 1 - x^{2}

\Rightarrow \sin (\arccos x) \sqrt{1 - x^{2}}

Now we can calculate the value of the original expression:

\sin (\arcsin x + \arccos x) =

= \sin (\arcsin x) \cdot \cos (\arccos x) + \cos (\arcsin x) \cdot \sin (\arccos y) =

= x \cdot x + \sqrt{1 - x^{2}} \cdot \sqrt{1 - x^{2}} =

= x^{2} + 1 - x^{2} = 1

How to simplify sin(arcsinx + arccosx)?

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