How do you solve $sin (arcsin (\frac{3}{5}) + arccos (\frac{3}{5}))$ $sin(arcsin(3/5) + arccos(3/5))$ ?

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How do you solve $sin (arcsin (\frac{3}{5}) + arccos (\frac{3}{5}))$ $sin(arcsin(3/5) + arccos(3/5))$ ?

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Answer 1 · 2015-05-15T23:07:47

Picture a right angled triangle with sides 3, 4 and 5 (since $3^{2} + 4^{2} = 5^{2}$ $3^2 + 4^2 = 5^2$ ). Call the smallest angle $α$ $alpha$ and the next smallest $β$ $beta$ .

$sin α = \frac{3}{5}$ $sin alpha = 3/5$ , being the length of the opposite side divided by the length of the hypotenuse.

$cos β = \frac{3}{5}$ $cos beta = 3/5$ , being the length of the adjacent side divided by the length of the hypotenuse.

So $arcsin (\frac{3}{5}) + arccos (\frac{3}{5}) = α + β = \frac{π}{2}$ $arcsin(3/5)+arccos(3/5) = alpha + beta = pi/2$ (or $90^{o}$ $90^o$ ) since together with the right angle, the internal angles of the triangle must add up to $π$ $pi$ ( $180^{o}$ $180^o$ ).

$sin (\frac{π}{2}) = 1$ $sin (pi/2) = 1$ .

So $sin (arcsin (\frac{3}{5}) + arccos (\frac{3}{5})) = 1$ $sin(arcsin(3/5)+arccos(3/5)) = 1$ .

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How do you solve $sin (arcsin (\frac{3}{5}) + arccos (\frac{3}{5}))$ $sin(arcsin(3/5) + arccos(3/5))$ ?

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