What is the value of sin(arccos(4/5)+arccos(5/13)) ?

The value of sin(arccos(4/5)+arccos(5/13)) is 63/65

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sin(arccos(4/5)+arccos(5/13))

sin (arccos (\frac{4}{5}) + arccos (\frac{5}{13}))

\frac{63}{65}

Solution steps

sin (arccos (\frac{4}{5}) + arccos (\frac{5}{13}))

Rewrite using trig identities:

sin (arccos (\frac{4}{5})) cos (arccos (\frac{5}{13})) + cos (arccos (\frac{4}{5})) sin (arccos (\frac{5}{13}))

= sin (arccos (\frac{4}{5})) cos (arccos (\frac{5}{13})) + cos (arccos (\frac{4}{5})) sin (arccos (\frac{5}{13}))

Rewrite using trig identities:

sin (arccos (\frac{4}{5})) = \frac{3}{5}

Rewrite using trig identities:

cos (arccos (\frac{5}{13})) = \frac{5}{13}

Rewrite using trig identities:

cos (arccos (\frac{4}{5})) = \frac{4}{5}

Rewrite using trig identities:

sin (arccos (\frac{5}{13})) = \frac{12}{13}

= \frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13}

Simplify

\frac{3}{5} \cdot \frac{5}{13} + \frac{4}{5} \cdot \frac{12}{13} : \frac{63}{65}

= \frac{63}{65}

40 sin (4 0^{\circ})

e^{1} sin (1)

4 + 3 cos (- \frac{π}{2})

4 + cot (\frac{π}{3}) - 2 csc (\frac{π}{3})

arctan (0.86)

What is the value of sin(arccos(4/5)+arccos(5/13)) ?
The value of sin(arccos(4/5)+arccos(5/13)) is 63/65