📏 trigonometry

1. Let's state the problem: We are given two right triangles with angles $61^\circ 30'$ and $56^\circ 20'$, and a vertical segment (flagpole) measuring 20 feet. We want to analyze

1. State the problem: Solve the equation $\sin 2x - \sin x = 0$. 2. Use the double-angle formula: $\sin 2x = 2 \sin x \cos x$, so the equation becomes:

1. The problem is to solve the equation $$2 \sin x \cos x (\cos^2 x - \sin^2 x) = \frac{1}{2}$$ for values of $x$. 2. Recall trigonometric identities:

1. We are asked to solve the equation $$2\sin x \cos x (\cos^2 x - \sin^2 x) = \frac{1}{2}.$$\n\n2. Recognize that $$2 \sin x \cos x = \sin 2x$$ and $$\cos^2 x - \sin^2 x = \cos 2x

1. Stating the problem: Solve the equation $$8 \sin^2(x) \cos^2(x) = 1$$. 2. Use the double-angle identity for sine: $$\sin(2x) = 2 \sin(x) \cos(x)$$, so $$\sin^2(2x) = 4 \sin^2(x)

1. Stating the problem: Solve the trigonometric equation $$\sin 2x \cos 2x - 2 \sin 2x = 0$$. 2. Factor the equation:

1. State the problem: Simplify the expression $$\sin(2x) \cdot \cos(2x) - 2 \sin(2x)$$. 2. Factor out the common term $$\sin(2x)$$:

1. **Find the arc length of a unit circle corresponding to the central angle measuring 60°.** The arc length $s$ on a circle is given by

1. **State the problem:** We are given a triangle with sides $a=4$, $b=6$, and angle $\alpha = 17^\circ$ opposite side $a$. We need to find the two possible sets of solutions for s

1. **State the problem:** Solve the equation $$\cos\left(\frac{x}{2}-1\right) = \cos^2\left(1 - \frac{x}{2}\right).$$ 2. **Rewrite the equation:** Let $$y = \frac{x}{2} - 1.$$ Then

1. We are given the equation $$\cos\left(\frac{\pi}{5} - \frac{1}{2}x\right) = -\frac{\sqrt{2}}{2}$$. We need to solve for $x$. 2. Recall that $$\cos(\theta) = -\frac{\sqrt{2}}{2}$

1. **Problem:** A balloon is 2500 ft above a lake. The angles of depression to the two opposite shores are 43° and 27°. We need to find the width of the lake. 2. **Setup:** Let the

1. For (h): Simplify the expression inside arcsin: $$\frac{3}{x} - \frac{3}{x} = 0.$$

1. The problem asks to convert 3\pi radians into grads. \ngrads and radians relate by the conversion factor: $$1 \text{ radian} = \frac{200}{\pi} \text{ grads}.$$\n\n2. To convert,

1. Convert 3\pi radians to grads. Since $$1\text{ radian} = \frac{200}{\pi} \text{ grads}$$, multiply: $$3\pi \times \frac{200}{\pi} = 600 \text{ grads}$$.

1. **Convert $\pi$ radians to grads.** Grads and radians are related by $$200 \text{ grads} = \pi \text{ radians}$$

1. Find $\tan\left(-\frac{2\pi}{3}\right)$. Recall that $\tan(\theta) = \frac{\sin \theta}{\cos \theta}$.

1. **Find \( \tan\left(-\frac{2\pi}{3}\right) \)** Step 1. Recognize the angle \( -\frac{2\pi}{3} \) is negative. Add \( 2\pi \) to find a positive coterminal angle:

1. The problem is to find the exact value of $\cot\left(-\frac{5\pi}{12}\right)$. 2. Recall that $\cot \theta = \frac{\cos \theta}{\sin \theta}$ and that cotangent is an odd functi

1. We'll start with a common Grade 10 trigonometry problem: Find the value of angle $\theta$ given $\sin \theta = 0.6$. 2. To find $\theta$, we use the inverse sine function: $$\th

1. Find the exact value of $\sin 45^\circ$. 2. Solve for $x$ if $\cos x = \frac{1}{2}$ and $0^\circ \leq x < 360^\circ$.