📏 trigonometry

1. Problem a: Find the new coordinates of point P after the transformation from $y=\tan x$ to $y=\tan 5x$. - Given original point $P=(45^\circ,1)$ on $y = \tan x$.

1. Stating the problem: Prove the identity $$\csc 2A = \frac{1}{2}(\sec A \csc A)$$. 2. Recall the double angle identity for cosecant: $$\csc 2A = \frac{1}{\sin 2A}$$.

1. **State the problem:** We are given a trigonometric function of the form $$g(x) = a \sin(bx + c) + d$$. We know two points: the function crosses its midline at (1.5, 1.5) and ha

1. The problem is to convert the angle \(\theta = 310^\circ\) from degrees to radians. 2. Recall the conversion formula: \(\text{radians} = \text{degrees} \times \frac{\pi}{180}\).

1. **State the problem:** We want to find the formula for a trigonometric function of the form $$h(x) = a \cos(bx + c) + d$$

1. We are given a vertical pole casting a shadow of 1.8 meters when the sun's elevation angle is 63.5 degrees. We need to find the height of the pole. 2. The problem can be solved

1. **Problem Statement:** Two signposts A and B are 11 km apart on a straight highway. An airplane is flying above the highway. The angles of depression from the airplane to A and

1. The problem is to understand the terms 'cahsign' which likely refers to 'cosine, adjacent, hypotenuse sign,' a mnemonic for trigonometric ratios. 2. The cosine of an angle in a

1. The problem is to find the value of $ (1 + \tan A)(1 + \tan B)$ where $A = 40^\circ$ and $B = 5^\circ$. 2. Substitute the given values:

1. The problem involves two expressions:\n$$\frac{1}{1} + \frac{1}{\text{something}} = 1$$\nand\n$$\frac{1 + \sin^2 A}{1 + \csc^2 A}$$\nWe need to analyze and simplify these.\n\n2.

1. **Problem statement:** Given $\sin \alpha = \frac{3}{5}$ with $\alpha$ in the first quadrant, find: - 5.3.1 $\tan \alpha$

1. Problem 5.1: Simplify $$\sin(90^\circ - x) \cdot \cos(180^\circ + x) + \tan x \cdot \cos x \cdot \sin(x - 180^\circ)$$ Step 1: Use angle identities:

1. The problem states that Pushkar observes the top of a pole $23^3$ m high with an angle of elevation of $30^\circ$, and the distance between Pushkar and the pole is 66 m. 2. Firs

1. **State the problem:** Pushkar observes the top of a pole that is $23^3$ meters high. The angle of elevation from Pushkar to the top of the pole is $30^\circ$, and the horizonta

1. დავიწყოთ პირველით: თუ $\sin x = \frac{\sqrt{2}}{2}$ და $x\in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$, მაშინ $x$ არის ის კუთხე, რომლის სინი არის $\frac{\sqrt{2}}{2}$. პოპულარ

1. Given $\sin x = \frac{\sqrt{2}}{2}$ and $x \in [-\frac{\pi}{2}, \frac{\pi}{2}]$, recognize that $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$. Since $\frac{\pi}{4}$ is within the in

1. **State the problem:** Show that $$\tan\left(\frac{45^\circ + \theta}{2}\right) \cdot \tan\left(\frac{45^\circ - \theta}{2}\right) = \frac{\sqrt{2} \cos \theta - 1}{\sqrt{2} \co

1. **Resolve a equação trigonométrica** $5\sin x - 6\cos x - 5 = 0$.\n Podemos reorganizar para isolar as funções trigonométricas:\n

1. State the problem: Express $3\cos \theta - 4\sin \theta$ in the form $R\cos(\theta - \alpha)$ and solve the equation $3\cos \theta - 4\sin \theta = 0$ for $0 \leq \theta \leq 2\

1. Let's start by understanding the quadrants. The first quadrant is where angles measure from $0^\circ$ to $90^\circ$, and the fourth quadrant is where angles measure from $270^\c

1. We are given that \(\tan(x) = \csc(x) - \sin(x)\) and need to prove that \(\tan(x^2)\left(\frac{x}{2}\right) = -2 + \sqrt{5}\).\n\n2. Start by expressing \(\csc(x)\) in terms of