Solve 6cos270^circ+10sin0^circ-4tan180^circ+5cos360^circ

$$6 \cos 270 ^ { \circ } + 10 \sin 0 ^ { \circ } - 4 \tan 180 ^ { \circ } + 5 \cos 360 ^ { \circ }$$

Evaluate

$$5$$

Solution Steps

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

$$6\times 0+10\sin(0)-4\tan(180)+5\cos(360)$$

Multiply 6 and 0 to get 0.

$$0+10\sin(0)-4\tan(180)+5\cos(360)$$

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

$$0+10\times 0-4\tan(180)+5\cos(360)$$

Multiply 10 and 0 to get 0.

$$0+0-4\tan(180)+5\cos(360)$$

Add 0 and 0 to get 0.

$$0-4\tan(180)+5\cos(360)$$

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

$$0-4\times 0+5\cos(360)$$

Multiply 4 and 0 to get 0.

$$0-0+5\cos(360)$$

Subtracting 0 from itself leaves 0.

$$0+5\cos(360)$$

Get the value of \cos(360) from trigonometric values table.

$$0+5\times 1$$

Multiply 5 and 1 to get 5.

$$0+5$$

Add 0 and 5 to get 5.

$$5$$