Question

cot 65degrees-cos25degrees/sin25degrees what's the exact value?

Answer 1

Answer on Question #36950 – Math – Trigonometry

cot 65 degrees-cos 25 degrees/sin 25 degrees what's the exact value?

Solution

$$\cot(65{}^\circ) - \frac{\cos(25{}^\circ)}{\sin(25{}^\circ)} = \frac{\cos(65{}^\circ)}{\sin(65{}^\circ)} - \frac{\cos(25{}^\circ)}{\sin(25{}^\circ)} = \frac{\cos(65{}^\circ) * \sin(25{}^\circ) - \cos(25{}^\circ) * \sin(65{}^\circ)}{\sin(65{}^\circ) * \sin(25{}^\circ)} =$$

Apply formulae

$$\cos(\alpha) * \sin(\beta) = \frac{1}{2}(\sin(\beta - \alpha) + \sin(\beta + \alpha))$$ $$\sin(\alpha) * \sin(\beta) = \frac{1}{2}(\cos(\alpha - \beta) - \cos(\beta + \alpha))$$

and calculate

$$\cos(65{}^\circ) * \sin(25{}^\circ) = \frac{1}{2}(\sin(25{}^\circ - 65{}^\circ) + \sin 90{}^\circ)) = \frac{1}{2}(1 - \sin 40{}^\circ)$$ $$\cos(25{}^\circ) * \sin(65{}^\circ) = \frac{1}{2}(\sin(40{}^\circ) + \sin 90{}^\circ) = \frac{1}{2}(\sin 40{}^\circ + 1)$$ $$\sin(65{}^\circ) * \sin(25{}^\circ) = \frac{1}{2}(\cos(65{}^\circ - 25{}^\circ) - \cos(25{}^\circ + 65{}^\circ)) = \frac{1}{2}(\cos(40{}^\circ) - 0) = \frac{1}{2} * \cos(40{}^\circ)$$

So,

$$\cot(65{}^\circ) - \frac{\cos(25{}^\circ)}{\sin(25{}^\circ)} = \frac{\cos(65{}^\circ)}{\sin(65{}^\circ)} - \frac{\cos(25{}^\circ)}{\sin(25{}^\circ)} = \frac{\frac{1}{2}(1 - \sin 40{}^\circ) - \frac{1}{2}(\sin 40{}^\circ + 1)}{\frac{1}{2} * \cos(40{}^\circ)} = -\tan(40{}^\circ) \approx -0.839$$