What is the value of sin(42)cos(48)+cos(42)sin(48) ?

The value of sin(42)cos(48)+cos(42)sin(48) is 1

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sin(42)cos(48)+cos(42)sin(48)

sin (4 2^{\circ}) cos (4 8^{\circ}) + cos (4 2^{\circ}) sin (4 8^{\circ})

1

Solution steps

sin (4 2^{\circ}) cos (4 8^{\circ}) + cos (4 2^{\circ}) sin (4 8^{\circ})

Rewrite using trig identities:

sin (9 0^{\circ})

sin (4 2^{\circ}) cos (4 8^{\circ}) + cos (4 2^{\circ}) sin (4 8^{\circ})

Use the Angle Sum identity:

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

= sin (4 2^{\circ} + 4 8^{\circ})

Simplify

= sin (9 0^{\circ})

= sin (9 0^{\circ})

Use the following trivial identity:

sin (9 0^{\circ}) = 1

sin (9 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= 1

cos (arcsin (\frac{5}{13}) + arctan (\frac{4}{3}))

cot (\frac{5}{12})

1 + sin (\frac{7 π}{4})

sin (- \frac{π}{6} - 2 ln (3 i))

6 + 3 cos (\frac{π}{2})

What is the value of sin(42)cos(48)+cos(42)sin(48) ?
The value of sin(42)cos(48)+cos(42)sin(48) is 1