Solve 5sin90^circ+2cos0^circ-2sin270^circ+10cos180^circ

$$5 \sin 90 ^ { \circ } + 2 \cos 0 ^ { \circ } - 2 \sin 270 ^ { \circ } + 10 \cos 180 ^ { \circ }$$

Evaluate

$$-1$$

Solution Steps

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

$$5\times 1+2\cos(0)-2\sin(270)+10\cos(180)$$

Multiply 5 and 1 to get 5.

$$5+2\cos(0)-2\sin(270)+10\cos(180)$$

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

$$5+2\times 1-2\sin(270)+10\cos(180)$$

Multiply 2 and 1 to get 2.

$$5+2-2\sin(270)+10\cos(180)$$

Add 5 and 2 to get 7.

$$7-2\sin(270)+10\cos(180)$$

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

$$7-2\left(-1\right)+10\cos(180)$$

Multiply 2 and -1 to get -2.

$$7-\left(-2\right)+10\cos(180)$$

The opposite of -2 is 2.

$$7+2+10\cos(180)$$

Add 7 and 2 to get 9.

$$9+10\cos(180)$$

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

$$9+10\left(-1\right)$$

Multiply 10 and -1 to get -10.

$$9-10$$

Subtract 10 from 9 to get -1.

$$-1$$