What is sin(180^0-β) ?

The solution to sin(180^0-β) is sin(1-β)

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simplify sin(180^0-β)

simplify

sin (18 0^{0} - β)

sin (1 - β)

Solution steps

sin (18 0^{0} - β)

Rewrite using trig identities

= sin (1) cos (β) - cos (1) sin (β)

Use the Angle Difference identity:

sin (s) cos (t) - cos (s) sin (t) = sin (s - t)

= sin (1 - β)

- cos (4 (\frac{π}{2}) + x)

sin^{2} (2 π x) + cos^{2} (2 π x)

\frac{2 d x}{sin ( x cos ( x ) )}

f (x) = 3 + cot (3) + cos (x) sec (3)

tan (arccos (\frac{x}{19}))