Question

college trigonometry. test #4
solve the problems on a sheet of paper. show your work, please. take a picture of your work and send it to me before the deadline. this quiz is based on the learning material \cotangent\ uploaded on dtl, so please read it carefully before working on the quiz.
in all problems below it’s assumed that the angle a is acute.
problem 1. given the following right - angled triangle, find sina, cosa, tana, cota.
problem 2. simplify 7·sinx·cotx
problem 3. simplify 8·tanx·cotx
problem 4. simplify cot²x + 1
problem 5. given sina = 12/13 find cos a, tan a and cota
problem 6. given cosa = √7/2 find 4sina+tana

Explanation:

Step1: Recall trigonometric ratios definitions

In a right - angled triangle, $\sin A=\frac{\text{opposite}}{\text{hypotenuse}}$, $\cos A=\frac{\text{adjacent}}{\text{hypotenuse}}$, $\tan A=\frac{\text{opposite}}{\text{adjacent}}$, $\cot A=\frac{\text{adjacent}}{\text{opposite}}$.

Step2: Identify sides for Problem 1

For the given right - angled triangle with opposite side to angle $A$ (side $BC$) $= 3$, adjacent side to angle $A$ (side $AC$) $= 5$ and hypotenuse (side $AB$) $=\sqrt{34}$.
$\sin A=\frac{3}{\sqrt{34}}=\frac{3\sqrt{34}}{34}$, $\cos A=\frac{5}{\sqrt{34}}=\frac{5\sqrt{34}}{34}$, $\tan A=\frac{3}{5}$, $\cot A=\frac{5}{3}$.

Step3: Simplify Problem 2

Recall that $\cot x=\frac{\cos x}{\sin x}$. Then $7\sin x\cot x = 7\sin x\times\frac{\cos x}{\sin x}=7\cos x$.

Step4: Simplify Problem 3

Since $\tan x=\frac{\sin x}{\cos x}$ and $\cot x=\frac{\cos x}{\sin x}$, then $8\tan x\cot x=8\times\frac{\sin x}{\cos x}\times\frac{\cos x}{\sin x}=8$.

Step5: Simplify Problem 4

Recall the Pythagorean identity $1 + \cot^{2}x=\csc^{2}x$.

Step6: Solve Problem 5

Given $\sin A=\frac{12}{13}$, using $\sin^{2}A+\cos^{2}A = 1$, we have $\cos A=\sqrt{1-\sin^{2}A}=\sqrt{1 - (\frac{12}{13})^{2}}=\sqrt{\frac{169 - 144}{169}}=\frac{5}{13}$. Then $\tan A=\frac{\sin A}{\cos A}=\frac{12}{5}$ and $\cot A=\frac{\cos A}{\sin A}=\frac{5}{12}$.

Step7: Solve Problem 6

Given $\cos A=\frac{\sqrt{7}}{2}$, using $\sin^{2}A+\cos^{2}A = 1$, $\sin A=\sqrt{1-\cos^{2}A}=\sqrt{1-\frac{7}{4}}$, but since $\frac{7}{4}>1$ this is a wrong - given value as for real - valued angles, $- 1\leqslant\cos A\leqslant1$.

Answer:

Problem 1: $\sin A=\frac{3\sqrt{34}}{34}$, $\cos A=\frac{5\sqrt{34}}{34}$, $\tan A=\frac{3}{5}$, $\cot A=\frac{5}{3}$
Problem 2: $7\cos x$
Problem 3: $8$
Problem 4: $\csc^{2}x$
Problem 5: $\cos A=\frac{5}{13}$, $\tan A=\frac{12}{5}$, $\cot A=\frac{5}{12}$
Problem 6: Given value of $\cos A=\frac{\sqrt{7}}{2}$ is incorrect for real - valued angles.