Solve 6sin(-90^circ)+3sin0^circ-8sin270^circ+12cos180^circ

$$6 \sin ( - 90 ^ { \circ } ) + 3 \sin 0 ^ { \circ } - 8 \sin 270 ^ { \circ } + 12 \cos 180 ^ { \circ }$$

Evaluate

$$-10$$

Solution Steps

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

$$6\left(-1\right)\sin(90)+3\sin(0)-8\sin(270)+12\cos(180)$$

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

$$6\left(-1\right)+3\sin(0)-8\sin(270)+12\cos(180)$$

Multiply 6 and -1 to get -6.

$$-6+3\sin(0)-8\sin(270)+12\cos(180)$$

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

$$-6+3\times 0-8\sin(270)+12\cos(180)$$

Multiply 3 and 0 to get 0.

$$-6+0-8\sin(270)+12\cos(180)$$

Add -6 and 0 to get -6.

$$-6-8\sin(270)+12\cos(180)$$

Get the value of \sin(270) from trigonometric values table.

$$-6-8\left(-1\right)+12\cos(180)$$

Multiply 8 and -1 to get -8.

$$-6-\left(-8\right)+12\cos(180)$$

The opposite of -8 is 8.

$$-6+8+12\cos(180)$$

Add -6 and 8 to get 2.

$$2+12\cos(180)$$

Get the value of \cos(180) from trigonometric values table.

$$2+12\left(-1\right)$$

Multiply 12 and -1 to get -12.

$$2-12$$

Subtract 12 from 2 to get -10.

$$-10$$