Question

problem 1. given the following right - angled triangle, find:
a) sina, cosa, tana, cota
b) sin b, cosb, tanb, cotb
problem 2. simplify
5sin²x + 6x + 5cos²x - 5
problem 3. simplify
7·cosx·tanx - 2sinx + tanx·cotx
problem 4. given cosa = \frac{2\sqrt{6}}{7} find sin a, tan a and cota
problem 5. given the triangle below find the value of the expression
10sina - 5 cosa + 3 tana + 6

Explanation:

Problem 1a

Step1: Recall trigonometric ratios

In a right - angled triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\), \(\cot\theta=\frac{\text{adjacent}}{\text{opposite}}\)
For angle \(A\), opposite side \(BC = 2\), adjacent side \(AC=1\), hypotenuse \(AB = \sqrt{5}\)

Step2: Calculate \(\sin A\)

\(\sin A=\frac{BC}{AB}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\)

Step3: Calculate \(\cos A\)

\(\cos A=\frac{AC}{AB}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)

Step4: Calculate \(\tan A\)

\(\tan A=\frac{BC}{AC}=\frac{2}{1} = 2\)

Step5: Calculate \(\cot A\)

\(\cot A=\frac{AC}{BC}=\frac{1}{2}\)

Problem 1b

Step1: Recall trigonometric ratios for angle \(B\)

For angle \(B\), opposite side \(AC = 1\), adjacent side \(BC = 2\), hypotenuse \(AB=\sqrt{5}\)

Step2: Calculate \(\sin B\)

\(\sin B=\frac{AC}{AB}=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}\)

Step3: Calculate \(\cos B\)

\(\cos B=\frac{BC}{AB}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}\)

Step4: Calculate \(\tan B\)

\(\tan B=\frac{AC}{BC}=\frac{1}{2}\)

Step5: Calculate \(\cot B\)

\(\cot B=\frac{BC}{AC}=2\)

Problem 2

Step1: Use the identity \(\sin^{2}x+\cos^{2}x = 1\)

\(5\sin^{2}x+6x + 5\cos^{2}x-5=5(\sin^{2}x+\cos^{2}x)+6x - 5\)

Step2: Substitute \(\sin^{2}x+\cos^{2}x = 1\)

\(5\times1+6x - 5=6x\)

Problem 3

Step1: Use the identities \(\tan x=\frac{\sin x}{\cos x}\) and \(\cot x=\frac{\cos x}{\sin x}\)

\(7\cos x\tan x-2\sin x+\tan x\cot x=7\cos x\times\frac{\sin x}{\cos x}-2\sin x+\frac{\sin x}{\cos x}\times\frac{\cos x}{\sin x}\)

Step2: Simplify the expression

\(7\sin x-2\sin x + 1=5\sin x+1\)

Problem 4

Step1: Use the identity \(\sin^{2}A+\cos^{2}A = 1\)

\(\sin A=\sqrt{1-\cos^{2}A}\), since \(\cos A=\frac{2\sqrt{6}}{7}\), then \(\cos^{2}A=\frac{24}{49}\)
\(\sin A=\sqrt{1-\frac{24}{49}}=\sqrt{\frac{49 - 24}{49}}=\sqrt{\frac{25}{49}}=\frac{5}{7}\)

Step2: Calculate \(\tan A\)

\(\tan A=\frac{\sin A}{\cos A}=\frac{\frac{5}{7}}{\frac{2\sqrt{6}}{7}}=\frac{5}{2\sqrt{6}}=\frac{5\sqrt{6}}{12}\)

Step3: Calculate \(\cot A\)

\(\cot A=\frac{\cos A}{\sin A}=\frac{\frac{2\sqrt{6}}{7}}{\frac{5}{7}}=\frac{2\sqrt{6}}{5}\)

Problem 5

Step1: Find the sides of the triangle

Using Pythagoras' theorem \(AC=\sqrt{AB^{2}-BC^{2}}=\sqrt{25 - 16}=3\)
For angle \(A\), \(\sin A=\frac{BC}{AB}=\frac{4}{5}\), \(\cos A=\frac{AC}{AB}=\frac{3}{5}\), \(\tan A=\frac{BC}{AC}=\frac{4}{3}\)

Step2: Substitute into the expression

\(10\sin A-5\cos A + 3\tan A+6=10\times\frac{4}{5}-5\times\frac{3}{5}+3\times\frac{4}{3}+6\)
\(=8 - 3+4 + 6=15\)

Answer:

• Problem 1a: \(\sin A=\frac{2\sqrt{5}}{5},\cos A=\frac{\sqrt{5}}{5},\tan A = 2,\cot A=\frac{1}{2}\)
• Problem 1b: \(\sin B=\frac{\sqrt{5}}{5},\cos B=\frac{2\sqrt{5}}{5},\tan B=\frac{1}{2},\cot B = 2\)
• Problem 2: \(6x\)
• Problem 3: \(5\sin x + 1\)
• Problem 4: \(\sin A=\frac{5}{7},\tan A=\frac{5\sqrt{6}}{12},\cot A=\frac{2\sqrt{6}}{5}\)
• Problem 5: \(15\)