Find the value of $sin ({cos}^{- 1} (\frac{3}{4}) - {tan}^{- 1 (\frac{1}{4})})$ $sin(cos^(-1)(3/4)-tan^(-1(1/4)))$ ?

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Find the value of $sin ({cos}^{- 1} (\frac{3}{4}) - {tan}^{- 1 (\frac{1}{4})})$ $sin(cos^(-1)(3/4)-tan^(-1(1/4)))$ ?

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Answer 1 · 2017-03-24T09:08:25

$sin ({cos}^{- 1} (\frac{3}{4}) - {tan}^{- 1} (\frac{1}{4})) = \frac{4 \sqrt{7} - 3}{4 \sqrt{17}}$ $sin(cos^(-1)(3/4)-tan^(-1)(1/4))=(4sqrt7-3)/(4sqrt17)$

Explanation:

Let ${cos}^{- 1} (\frac{3}{4}) = α$ $cos^(-1)(3/4)=alpha$ , then $cos α = \frac{3}{4}$ $cosalpha=3/4$

and $sin α = \sqrt{1 - {cos}^{2} α}$ $sinalpha=sqrt(1-cos^2alpha)$

= $\sqrt{1 - \frac{9}{16}} = \frac{\sqrt{7}}{4}$ $sqrt(1-9/16)=sqrt7/4$

let ${tan}^{- 1} (\frac{1}{4}) = β$ $tan^(-1)(1/4)=beta$ then we have $tan β = \frac{1}{4}$ $tanbeta=1/4$

and $cos β = \frac{1}{sec β} = \frac{1}{\sqrt{1 + {tan}^{2} β}} = \frac{1}{\sqrt{1 + \frac{1}{16}}} = \frac{4}{\sqrt{17}}$ $cosbeta=1/secbeta=1/sqrt(1+tan^2beta)=1/sqrt(1+1/16)=4/sqrt17$

and $sin β = tan β \times cos β = \frac{1}{4} \times \frac{4}{\sqrt{17}} = \frac{1}{\sqrt{17}}$ $sinbeta=tanbetaxxcosbeta=1/4xx4/sqrt17=1/sqrt17$

Hence $sin ({cos}^{- 1} (\frac{3}{4}) - {tan}^{- 1} (\frac{1}{4}))$ $sin(cos^(-1)(3/4)-tan^(-1)(1/4))$

= $sin (α - β) = sin α cos β - cos α sin β$ $sin(alpha-beta)=sinalphacosbeta-cosalphasinbeta$

= $\frac{\sqrt{7}}{4} \times \frac{4}{\sqrt{17}} - \frac{3}{4} \times \frac{1}{\sqrt{17}}$ $sqrt7/4xx4/sqrt17-3/4xx1/sqrt17$

= $\frac{4 \sqrt{7} - 3}{4 \sqrt{17}}$ $(4sqrt7-3)/(4sqrt17)$

Answer 2 · 2017-03-24T09:49:00

$\frac{\sqrt{17} (4 \sqrt{7} - 3)}{68}$ $(sqrt17(4sqrt7-3))/(68)$

Explanation:

The given expression
$sin ({cos}^{- 1} (\frac{3}{4}) - {tan}^{- 1} (\frac{1}{4}))$ $sin (cos^-1(3/4) - tan^-1(1/4))$
looks like $sin (A - B)$ $sin (A-B)$
We also know that
$sin (A - B) = sin A cos B - cos A sin B$ $sin(A-B)=sinAcosB-cosAsinB$ ....(1)
where
$A = {cos}^{- 1} (\frac{3}{4})$ $A=cos^-1(3/4)$ .....(2)
and
$B = {tan}^{- 1} (\frac{1}{4})$ $B= tan^-1(1/4)$ ......(3)

To calculate $sin A and cos A$ $sin A and cos A$ we make use of (2)
From (2)
$cos A = cos [{cos}^{- 1} (\frac{3}{4})]$ $cos A=cos [cos^-1(3/4)]$
$cos A = \frac{3}{4}$ $cos A=3/4$ .....(4)
Using the identity
${sin}^{2} A + {cos}^{2} A = 1$ $sin^2A+cos^2A=1$
$sin A = \sqrt{1 - {cos}^{2} A}$ $sinA=sqrt(1-cos^2A)$
$\Rightarrow sin A = \sqrt{1 - {(\frac{3}{4})}^{2}}$ $=>sinA=sqrt(1-(3/4)^2)$
$\Rightarrow sin A = \sqrt{1 - \frac{9}{16}}$ $=>sinA=sqrt(1-9/16)$
$\Rightarrow sin A = \frac{\sqrt{7}}{4}$ $=>sinA=sqrt7/4$ ......(5)

To calculate $sin B and cos B$ $sin B and cos B$ we make use of (3)
From (3)
$tan B = tan [{tan}^{- 1} (\frac{1}{4})]$ $tanB= tan[tan^-1(1/4)]$
$\Rightarrow tan B = \frac{1}{4}$ $=>tanB= 1/4$ .....(6)
we know that
$cos B = \frac{1}{sec B}$ $cosB=1/(secB)$
$\Rightarrow cos B = \frac{1}{\sqrt{1 + {tan}^{2} B}}$ $=>cosB=1/sqrt(1+tan^2B)$
Inserting value of $tan B$ $tanB$ we get
$\Rightarrow cos B = \frac{1}{\sqrt{1 + {(\frac{1}{4})}^{2}}}$ $=>cosB=1/sqrt(1+(1/4)^2)$
$\Rightarrow cos B = \frac{1}{\sqrt{1 + \frac{1}{16}}}$ $=>cosB=1/sqrt(1+1/16)$
$\Rightarrow cos B = \frac{4}{\sqrt{17}}$ $=>cosB=4/sqrt(17)$ ......(7)

We know that
$sin B = tan B \times cos B$ $sinB=tanBxxcosB$
Using (6) and (7) we get
$sin B = \frac{1}{4} \times \frac{4}{\sqrt{17}}$ $sinB=1/4xx4/sqrt17$
$sin B = \frac{1}{\sqrt{17}}$ $sinB=1/sqrt17$ .....(8)

Inserting calculated values in (1) we get
$sin (A - B) = \frac{\sqrt{7}}{4} \times \frac{4}{\sqrt{17}} - \frac{3}{4} \times \frac{1}{\sqrt{17}}$ $sin(A-B)=sqrt7/4xx4/sqrt17-3/4xx1/sqrt17$
$sin (A - B) = \frac{4 \sqrt{7} - 3}{4 \sqrt{17}}$ $sin(A-B)=(4sqrt7-3)/(4sqrt17)$
Rationalizing the denominator we get
$sin (A - B) = \frac{\sqrt{17} (4 \sqrt{7} - 3)}{68}$ $sin(A-B)=(sqrt17(4sqrt7-3))/(68)$

.-.-.-.-.-.-.-.-.-.-

Alternatively after (4)
In a right angled triangle
$cos A = \frac{3}{4} \equiv \frac{base}{hypotenuse}$ $cosA=3/4-="base"/"hypotenuse"$
and
$perpendicular = \sqrt{{hypotenuse}^{2} - {base}^{2}}$ $"perpendicular"=sqrt("hypotenuse"^2-"base"^2)$
$:. "perpendicular"=sqrt(4^2-3^2)$
$=> "perpendicular"=sqrt(7)$
This gives us
$sinA-="perpendicular"/"hypotenuse"=sqrt7/4$
This is same as (5)

Similar steps can be followed after (6) to calculate $sinB and cos B$ that $tanB=1/4-="perpendicular"/"base"$
and
$"hypotenuse"=sqrt("perpendicular"^2+"base"^2)$

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