What is the value of sin(arcsin(1)+arccos(1)) ?

The value of sin(arcsin(1)+arccos(1)) is 1

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sin(arcsin(1)+arccos(1))

sin (arcsin (1) + arccos (1))

1

Solution steps

sin (arcsin (1) + arccos (1))

Rewrite using trig identities:

sin (arcsin (1)) cos (arccos (1)) + cos (arcsin (1)) sin (arccos (1))

sin (arcsin (1) + arccos (1))

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (arcsin (1)) cos (arccos (1)) + cos (arcsin (1)) sin (arccos (1))

= sin (arcsin (1)) cos (arccos (1)) + cos (arcsin (1)) sin (arccos (1))

Rewrite using trig identities:

sin (arcsin (1)) = 1

Use the following identity:

sin (arcsin (x)) = x

= 1

Rewrite using trig identities:

cos (arccos (1)) = 1

Use the following identity:

cos (arccos (x)) = x

= 1

Rewrite using trig identities:

cos (arcsin (1)) = 0

cos (arcsin (1))

Rewrite using trig identities:

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

Use the following identity:

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

= 1 - 1^{2}

= 1 - 1^{2}

Simplify

= 0

Rewrite using trig identities:

sin (arccos (1)) = 0

sin (arccos (1))

Rewrite using trig identities:

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

Use the following identity:

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

= 1 - 1^{2}

= 1 - 1^{2}

Simplify

= 0

= 1 \cdot 1 + 0 \cdot 0

Simplify

= 1

40 \cdot \frac{sin ( 13 5 ^{\circ} )}{sin ( 10 5 ^{\circ} )}

8 sin (29 0^{\circ}) + 6 sin (6 0^{\circ})

arccos (cos (\frac{π}{12}))

\frac{10}{sin ( 3 6 ^{\circ} )}

4 (sin (15 0^{\circ})) (cos (21 0^{\circ})) + 3 (tan (24 0^{\circ}))

What is the value of sin(arcsin(1)+arccos(1)) ?
The value of sin(arcsin(1)+arccos(1)) is 1