Solve 6sin(-90^circ)+3sin0^circ-8sin2)0^circ+12cos180^circ

$$6 \sin ( - 90 ^ { \circ } ) + 3 \sin 0 ^ { \circ } - 8 \sin 2 ) 0 ^ { \circ } + 12 \cos 180 ^ { \circ }$$

Evaluate

$$-6.27919597362000776$$

Solution Steps

Evaluate trigonometric functions in the problem

$$6 \sin(-90) + 3 \sin(0) - 8 \cdot 0.03489949670250097$$

Use the property \sin(-x)=-\sin(x).

$$6\left(-1\right)\sin(90)+3\sin(0)-8\times 0.03489949670250097$$

Get the value of \sin(90) from trigonometric values table.

$$6\left(-1\right)+3\sin(0)-8\times 0.03489949670250097$$

Multiply 6 and -1 to get -6.

$$-6+3\sin(0)-8\times 0.03489949670250097$$

Get the value of \sin(0) from trigonometric values table.

$$-6+3\times 0-8\times 0.03489949670250097$$

Multiply 3 and 0 to get 0.

$$-6+0-8\times 0.03489949670250097$$

Add -6 and 0 to get -6.

$$-6-8\times 0.03489949670250097$$

Multiply 8 and 0.03489949670250097 to get 0.27919597362000776.

$$-6-0.27919597362000776$$

Subtract 0.27919597362000776 from -6 to get -6.27919597362000776.

$$-6.27919597362000776$$