How do you simplify $cos [{tan}^{- 1} (- 1) + {cos}^{- 1} (- \frac{4}{5})]$ $Cos [tan^-1(-1) + cos^-1(-4/5)]$ ?

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How do you simplify $cos [{tan}^{- 1} (- 1) + {cos}^{- 1} (- \frac{4}{5})]$ $Cos [tan^-1(-1) + cos^-1(-4/5)]$ ?

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Answer 1 · 2017-03-20T20:51:48

see below

Explanation:

Use the property $cos (A + B) = cos A cos B - sin A sin B$ $color(blue)(cos(A+B)=cosAcosB-sinAsinB$

For the ${tan}^{- 1} (- 1)$ $color(red)(tan^-1 (-1)$ we know that our triangle is in quadrant 4. Since ${tan}^{- 1} x$ $color(red)(tan^-1 x$ is restricted to quadrants 1 and 4 and since the argument is negative it means our triangle has to be in quadrant 4.

Hence, the side $o p p o s i t e$ $color(red)(opposite$ to $∠ θ = - 1$ $color(red)(angle theta = -1$ and $a d j a c e n t$ $color(red)(adjacent$ to $∠ θ = 1$ $color(red)(angle theta = 1$ and so the $h y p o t e n u s e = \sqrt{2}$ $color(red)(hypote n use=sqrt2$ . In this case $θ = 45^{\circ}$ $color(red)(theta=45^@$ but we don't need to know the value of $θ$ $color(red)(theta$ . Therefore,

$cos ({tan}^{- 1} (- 1)) = cos θ = \frac{a d j a c e n t}{h y p o t e n u s e} = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2}$ $color(blue)(cos(tan^-1(-1))=cos theta=(adjacent)/(hypote n use)=1/sqrt2 = sqrt2/2$

$sin ({tan}^{- 1} (- 1)) = sin θ = \frac{o p p o s i t e}{h y p o t e n u s e} = - \frac{1}{\sqrt{2}} = - \frac{\sqrt{2}}{2}$ $color(blue)(sin(tan^-1(-1))=sin theta=(opposite)/(hypote n use)=-1/sqrt2 = -sqrt2/2$

For the ${cos}^{- 1} (- \frac{4}{5})$ $color(magenta)(cos^-1(-4/5)$ the triangle is in quadrant 2 since ${cos}^{- 1} x$ $color(magenta)(cos^-1 x)$ is restricted to quadrants 1 and 2 and since the argument is negative the triangle has to be in quadrant 2.

Hence, $a d j a c e n t$ $color(magenta)(adjacent$ to $∠ θ = - 4$ $color(magenta)(angle theta=-4$ and $h y p o t e n u s e = 5$ $color(magenta)(hypote n use = 5$ . Therefore by using pythagorean theorem the $o p p o s i t e$ $color(magenta)(opposite$ to $∠ θ = 3$ $color(magenta)(angle theta=3$

$(cos ({cos}^{- 1} (- \frac{4}{5})) = cos θ = \frac{a d j a c e n t}{h y p o t e n u s e} = - \frac{4}{5}$ $color(blue)((cos(cos^-1(-4/5))=cos theta = (adjacent)/(hypote n use)=-4/5$

$sin ({cos}^{- 1} (- \frac{4}{5}) = sin θ = \frac{o p p o s i t e}{h y p o t e n u s e} = \frac{3}{5}$ $color(blue)(sin(cos^-1(-4/5)=sin theta=(opposite)/(hypote n use) = 3/5$

Then using the property $cos (A + B) = cos A cos B - sin A sin B$ $color(orange)(cos(A+B)=cosAcosB-sinAsinB$ where $A = {tan}^{- 1} (- 1) and B = {cos}^{- 1} (- \frac{4}{5})$ $color(orange)(A=tan^-1 (-1) and B=cos^-1(-4/5)$ we have

$cos ({tan}^{- 1} (- 1) + {cos}^{- 1} (- \frac{4}{5})) = cos ({tan}^{- 1} (- 1)) cos ({cos}^{- 1} (- \frac{4}{5})) - sin ({tan}^{- 1} (- 1)) sin ({cos}^{- 1} (- \frac{4}{5}))$ $cos(tan^-1(-1)+cos^-1(-4/5))=color(blue)(cos(tan^-1(-1))cos(cos^-1(-4/5))-sin(tan^-1(-1)) sin(cos^-1(-4/5))$

$= \frac{\sqrt{2}}{2} \cdot \frac{- 4}{5} - \frac{- \sqrt{2}}{2} \cdot \frac{3}{5}$ $color(blue)(=sqrt2/2 * (-4)/5-(-sqrt2)/2*3/5$

$= \frac{- 4 \sqrt{2}}{10} + \frac{3 \sqrt{2}}{10}$ $color(blue)(=(-4sqrt2)/10)+color(blue)((3sqrt2)/10$

$color(blue)( :. = -sqrt2/10$

Answer 2 · 2017-03-21T03:33:51

$LHS=cos(tan^-1(-1)+cos^-1(-4/5))$

$=cos(tan^-1tan(-pi/4)+cos^-1(-4/5))$

$=cos((-pi/4)+cos^-1(-4/5))$

$=cos(-pi/4)cos(cos^-1(-4/5))-sin(-pi/4)sin(cos^-1(-4/5))$

$=cos(pi/4)(-4/5))+sin(pi/4)sin(sin^-1sqrt(1-(-4/5)^2))$

$=1/sqrt2xx(-4/5)+1/sqrt2xx3/5$

$=(-4+3)/(5sqrt2)$

$=-1/(5sqrt2)=-sqrt2/10$

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