What is the value of 3+2sin(-pi/2) ?

The value of 3+2sin(-pi/2) is 1

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3+2sin(-pi/2)

3 + 2 sin (- \frac{π}{2})

1

Solution steps

3 + 2 sin (- \frac{π}{2})

Rewrite using trig identities:

sin (- \frac{π}{2}) = - 1

sin (- \frac{π}{2})

Use the following property:

sin (- x) = - sin (x)

sin (- \frac{π}{2}) = - sin (\frac{π}{2})

= - sin (\frac{π}{2})

Use the following trivial identity:

sin (\frac{π}{2}) = 1

sin (\frac{π}{2})

sin (x)

periodicity table with

2 πn

cycle:

x 0 \frac{π}{6} \frac{π}{4} \frac{π}{3} \frac{π}{2} \frac{2 π}{3} \frac{3 π}{4} \frac{5 π}{6} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x π \frac{7 π}{6} \frac{5 π}{4} \frac{4 π}{3} \frac{3 π}{2} \frac{5 π}{3} \frac{7 π}{4} \frac{11 π}{6} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= 1

= - 1

= 3 + 2 (- 1)

Simplify

= 1

tan (24 0^{\circ}) - tan (21 0^{\circ})

arctan (\frac{3}{1})

9 cos (\frac{π}{4})

arctan (\frac{3}{- 8})

arctan (1.87)