Question

Use the given information to calculate the exact value of the expression please help trig?

5.cosx=-√429/25 cosy=3/5 sinx=14/25 siny=4/5
a. find cos(x+y) b. find tan(x-y) c. find sin(x-y)
6. cos a=-12/16 pi/2 ≤ a ≤ pi and sin b= 6/8 pi ≤ 6 ≤ -3pi/2
a. find sin(a-b) b. find cot(a+b) c. find cos(a-b)

Answer 1

Question#29145

Use the given information to calculate the exact value of the expression please help trig?

5.cosx=-√429/25 cosy=3/5 sinx=14/25 siny=4/5

a. find cos(x+y) b. find tan(x-y) c. find sin(x-y)

6. cos a=-12/16 pi/2 ≤ a ≤ pi and sin b= 6/8 pi ≤ 6 ≤ -3pi/2

a. find sin(a-b) b. find cot(a+b) c. find cos(a-b)

Solution:

5.

a. $\cos(x + y) = \cos x \cos y - \sin x \sin y = -\frac{\sqrt{429} \cdot 3}{25} \cdot \frac{14}{25} \cdot \frac{4}{5} = -\frac{3\sqrt{429} + 56}{125}$

b. $\tan x = \frac{\sin x}{\cos x} = \frac{\frac{14}{25}}{\frac{\sqrt{429}}{25}} = -\frac{14 \cdot 25}{25 \cdot \sqrt{429}} = -\frac{14}{\sqrt{429}}$

$$\tan y = \frac{\sin y}{\cos y} = \frac{5}{5} = \frac{4 \cdot 5}{5 \cdot 3} = \frac{4}{3}$$ $$\tan(x - y) = \frac{\tan x - \tan y}{1 + \tan x + \tan y} = \frac{-\frac{14}{\sqrt{429}} - \frac{4}{3}}{1 + \left(-\frac{14}{\sqrt{429}}\right) \cdot \frac{4}{3}} = \frac{\frac{42 + 4\sqrt{429}}{3\sqrt{429}}}{\frac{3\sqrt{429} - 56}{3\sqrt{429}}} = \frac{42 + 4\sqrt{429}}{56 - 3\sqrt{429}}$$

c. $\sin(x - y) = \sin x \cos y - \cos x \sin y = \frac{14}{25} \cdot \frac{3}{5} - \left(-\frac{\sqrt{429}}{25}\right) \cdot \frac{4}{5} = \frac{42 + 4\sqrt{429}}{125}$

6. $\sin a = +\sqrt{1 - \cos^2 a} = \sqrt{1 - \left(-\frac{12^2}{16}\right)} = \sqrt{1 - \frac{144}{256}} = \sqrt{\frac{112}{256}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$

$$\cos b = -\sqrt{1 - \sin^2 b} = -\sqrt{1 - \left(\frac{6}{8}\right)^2} = -\sqrt{1 - \frac{36}{48}} = -\sqrt{\frac{28}{64}} = -\frac{\sqrt{7}}{4}$$

6.a

$$\sin(a - b) = \sin a \cos b - \cos a \sin b = \frac{\sqrt{7}}{4} \cdot \left(-\frac{\sqrt{7}}{4}\right) - \left(-\frac{12}{16}\right) \cdot \frac{6}{8} = \frac{-56 + 72}{16 \cdot 8} = \frac{16}{16 \cdot 8} = \frac{1}{8}$$

6.b

$$\cot a = \frac {\cos a}{\sin a} = \frac {- \frac {12}{16}}{\frac {\sqrt {7}}{4}} = - \frac {3}{\sqrt {7}}$$ $$\cot b = \frac {\cos b}{\sin b} = \frac {- \frac {\sqrt {7}}{4}}{\frac {6}{8}} = - \frac {\sqrt {7}}{3}$$ $$\cot (a + b) = \frac {\cot a \cot b - 1}{\cot a + \cot b} = \frac {- \frac {3}{\sqrt {7}} * \left(- \frac {\sqrt {7}}{3}\right) - 1}{- \frac {3}{\sqrt {7}} * \left(- \frac {\sqrt {7}}{3}\right)} = 0$$

6.c

$$\cos (a - b) = \cos a \cos b + \sin a \sin b = - \frac {12}{16} * \left(- \frac {\sqrt {7}}{4}\right) + \frac {\sqrt {7}}{4} * \frac {6}{8} = \frac {24\sqrt {7}}{16 + 4} \frac {3\sqrt {7}}{8}$$