Question

what is the exact value of 4cos(540degrees)+3tan(-405)degrees

Answer 1

Answer on Question#31412 – Math – Trigonometry

Condition of the problem:

What is the exact value of $4\cos(540{}^\circ) + 3\tg(-405{}^\circ)$?

Solution:

It is known that

$$540{}^\circ = 3\pi,$$ $$\cos(540{}^\circ) = \cos(3\pi) = -1.$$

It is known that

$$405{}^\circ = 360{}^\circ + 45{}^\circ,$$

Using next formulas to calculate the $\tg(-405{}^\circ)$:

$$\tg(x) = \frac{\sin(x)}{\cos(x)},$$ $$\sin(-x) = -\sin(x), \quad \cos(-x) = \cos(x),$$ $$\sin(x + y) = \sin(x)\cos(y) + \sin(y)\cos(x),$$ $$\cos(x + y) = \cos(x)\cos(y) - \sin(x)\sin(y),$$ $$\sin(360{}^\circ) = 0, \quad \cos(360{}^\circ) = 1,$$ $$\sin(45{}^\circ) = \frac{\sqrt{2}}{2}, \quad \cos(45{}^\circ) = \frac{\sqrt{2}}{2}.$$ $$\tg(-405{}^\circ) = \frac{\sin(-405{}^\circ)}{\cos(-405{}^\circ)} = \frac{-\sin(405{}^\circ)}{\cos(405{}^\circ)} = \frac{-\sin(360{}^\circ + 45{}^\circ)}{\cos(360{}^\circ + 45{}^\circ)} = \frac{-\sin(360{}^\circ)\cos(45{}^\circ) - \sin(45{}^\circ)\cos(360{}^\circ)}{\cos(360{}^\circ)\cos(45{}^\circ) - \sin(45{}^\circ)\sin(360{}^\circ)} = \frac{-\sin(45{}^\circ)}{\cos(45{}^\circ)} = \frac{-\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -1.$$

Conclusion:

$$4\cos(540{}^\circ) + 3\tg(-405{}^\circ) = 4 \cdot (-1) + 3 \cdot (-1) = -7.$$

Answer: -7.