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 = − 429 ⋅ 3 25 ⋅ 14 25 ⋅ 4 5 = − 3 429 + 56 125 \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} cos ( x + y ) = cos x cos y − sin x sin y = − 25 429 ⋅ 3 ⋅ 25 14 ⋅ 5 4 = − 125 3 429 + 56
b. tan x = sin x cos x = 14 25 429 25 = − 14 ⋅ 25 25 ⋅ 429 = − 14 429 \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 x = c o s x s i n x = 25 429 25 14 = − 25 ⋅ 429 14 ⋅ 25 = − 429 14
tan y = sin y cos y = 5 5 = 4 ⋅ 5 5 ⋅ 3 = 4 3 \tan y = \frac{\sin y}{\cos y} = \frac{5}{5} = \frac{4 \cdot 5}{5 \cdot 3} = \frac{4}{3} tan y = cos y sin y = 5 5 = 5 ⋅ 3 4 ⋅ 5 = 3 4 tan ( x − y ) = tan x − tan y 1 + tan x + tan y = − 14 429 − 4 3 1 + ( − 14 429 ) ⋅ 4 3 = 42 + 4 429 3 429 3 429 − 56 3 429 = 42 + 4 429 56 − 3 429 \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}} tan ( x − y ) = 1 + tan x + tan y tan x − tan y = 1 + ( − 429 14 ) ⋅ 3 4 − 429 14 − 3 4 = 3 429 3 429 − 56 3 429 42 + 4 429 = 56 − 3 429 42 + 4 429
c. sin ( x − y ) = sin x cos y − cos x sin y = 14 25 ⋅ 3 5 − ( − 429 25 ) ⋅ 4 5 = 42 + 4 429 125 \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} sin ( x − y ) = sin x cos y − cos x sin y = 25 14 ⋅ 5 3 − ( − 25 429 ) ⋅ 5 4 = 125 42 + 4 429
6. sin a = + 1 − cos 2 a = 1 − ( − 1 2 2 16 ) = 1 − 144 256 = 112 256 = 7 16 = 7 4 \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} sin a = + 1 − cos 2 a = 1 − ( − 16 1 2 2 ) = 1 − 256 144 = 256 112 = 16 7 = 4 7
cos b = − 1 − sin 2 b = − 1 − ( 6 8 ) 2 = − 1 − 36 48 = − 28 64 = − 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} cos b = − 1 − sin 2 b = − 1 − ( 8 6 ) 2 = − 1 − 48 36 = − 64 28 = − 4 7
6.a
sin ( a − b ) = sin a cos b − cos a sin b = 7 4 ⋅ ( − 7 4 ) − ( − 12 16 ) ⋅ 6 8 = − 56 + 72 16 ⋅ 8 = 16 16 ⋅ 8 = 1 8 \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} sin ( a − b ) = sin a cos b − cos a sin b = 4 7 ⋅ ( − 4 7 ) − ( − 16 12 ) ⋅ 8 6 = 16 ⋅ 8 − 56 + 72 = 16 ⋅ 8 16 = 8 1
6.b
cot a = cos a sin a = − 12 16 7 4 = − 3 7 \cot a = \frac {\cos a}{\sin a} = \frac {- \frac {12}{16}}{\frac {\sqrt {7}}{4}} = - \frac {3}{\sqrt {7}} cot a = sin a cos a = 4 7 − 16 12 = − 7 3 cot b = cos b sin b = − 7 4 6 8 = − 7 3 \cot b = \frac {\cos b}{\sin b} = \frac {- \frac {\sqrt {7}}{4}}{\frac {6}{8}} = - \frac {\sqrt {7}}{3} cot b = sin b cos b = 8 6 − 4 7 = − 3 7 cot ( a + b ) = cot a cot b − 1 cot a + cot b = − 3 7 ∗ ( − 7 3 ) − 1 − 3 7 ∗ ( − 7 3 ) = 0 \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 cot ( a + b ) = cot a + cot b cot a cot b − 1 = − 7 3 ∗ ( − 3 7 ) − 7 3 ∗ ( − 3 7 ) − 1 = 0
6.c
cos ( a − b ) = cos a cos b + sin a sin b = − 12 16 ∗ ( − 7 4 ) + 7 4 ∗ 6 8 = 24 7 16 + 4 3 7 8 \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} cos ( a − b ) = cos a cos b + sin a sin b = − 16 12 ∗ ( − 4 7 ) + 4 7 ∗ 8 6 = 16 + 4 24 7 8 3 7 
