Question

If tanx =3/4, π < x < 3 π /2, find the values of sin (x/2), cos (x/2) and tan (x/2)?

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Answer 1

$\begin{gathered}\begin{gathered}\bf \: Given - \begin{cases} &\sf{tanx \: = \: \dfrac{3}{4} } \\ &\sf{and \: \pi \: < x < \dfrac{3\pi}{2} } \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf \: To\:find - \begin{cases} &\sf{sin\dfrac{x}{2} } \\ &\sf{cos\dfrac{x}{2} } \\ &\sf{tan\dfrac{x}{2} }  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$(1). \: \boxed{ \bf{ {sec}^{2}x - {tan}^{2}x = 1}}$

$(2). \:\boxed{ \bf{cos2x = 1 -2 {sin}^{2}x}}$

$(3). \:\boxed{ \bf{cos2x = 2 {cos}^{2}x - 1 }}$

$\large\underline{\bold{Solution :- }}$

$\large \underline{\tt \:{ According \: to \: statement }}$

$\rm :\implies\:tanx = \dfrac{3}{4}$

We know, that

$\rm :\longmapsto\: {sec}^{2} x - {tan}^{2} x = 1$

$\rm :\longmapsto\: {sec}^{2} x - ({\dfrac{3}{4} )}^{2} = 1$

$\rm :\implies\: {sec}^{2} x = 1 + \dfrac{9}{16}$

$\rm :\longmapsto\: {sec}^{2} x = \dfrac{25}{16}$

$\rm :\implies\:secx \: = \pm \: \dfrac{5}{4}$

As,

$\rm :\longmapsto\:\pi \: < \: x \: < \: \dfrac{3\pi}{2}$

So,

• secx < 0

$\bf :\implies\:secx = - \dfrac{5}{4}$

$\bf:\implies\:cosx = - \dfrac{4}{5}$

Evaluation of sin (x/2)

• We know that

$\rm :\longmapsto\:cosx = 1 - 2 {sin}^{2} \dfrac{x}{2}$

$\rm :\longmapsto\: - \dfrac{4}{5} = 1 - 2 {sin}^{2} \dfrac{x}{2}$

$\rm :\longmapsto\:2 {sin}^{2} \dfrac{x}{2} = 1 + \dfrac{4}{5}$

$\rm :\longmapsto\:2 {sin}^{2} \dfrac{x}{2} = \dfrac{5 + 4}{5}$

$\rm :\longmapsto\: {sin}^{2} \dfrac{x}{2} = \dfrac{9}{10}$

$\rm :\implies\:sin\dfrac{x}{2} = \pm \: \dfrac{3}{ \sqrt{10} }$

Now,

$\rm :\longmapsto\:\pi \: < \: x \: < \: \dfrac{3\pi}{2}$

$\rm :\implies\:\dfrac{\pi}{2} < \dfrac{x}{2} < \dfrac{3\pi}{4}$

$\rm :\longmapsto\:\dfrac{x}{2} \: \in \: {2}^{nd} \: quadrant.$

$\bf\implies \:sin\dfrac{x}{2} > 0$

$\bf\implies \: \boxed{ \bf \: sin\dfrac{x}{2} = \dfrac{3}{ \sqrt{10} } }$

Evaluation of cos (x/2).

Now,

• We know that

$\rm :\longmapsto\:cosx = 2 {cos}^{2} \dfrac{x}{2} - 1$

$\rm :\longmapsto\: - \dfrac{4}{5} = {2cos}^{2} \dfrac{x}{2} - 1$

$\rm :\implies\:2 {cos}^{2} \dfrac{x}{2} = 1 - \dfrac{4}{5}$

$\rm :\longmapsto\:2 {cos}^{2} \dfrac{x}{2} = \dfrac{5 - 4}{5}$

$\rm :\longmapsto\: {cos}^{2} \dfrac{x}{2} = \dfrac{1}{5}$

$\rm :\implies\:cos\dfrac{x}{2} = \: \pm \: \dfrac{1}{ \sqrt{10} }$

$\rm :\longmapsto\:As \: \dfrac{x}{2} \: \in \: {2}^{nd} \: quadrant$

$\rm :\implies\:cos\dfrac{x}{2} < 0$

$\rm :\implies\: \boxed{ \bf \: cos\dfrac{x}{2} = - \dfrac{1}{ \sqrt{10} } }$

Now,

• we know that

$\rm :\longmapsto\:tan\dfrac{x}{2} = \dfrac{sin\dfrac{x}{2}}{cos\dfrac{x}{2}}$

$\rm :\implies\:tan\dfrac{x}{2} = \dfrac{\dfrac{3}{ \sqrt{10} } }{\dfrac{ - 1}{ \sqrt{10} } }$

$\rm :\implies\: \boxed{ \bf \: tan\dfrac{x}{2} = - 3}$

$\begin{gathered}\begin{gathered}\bf \: Hence - \begin{cases} &\sf{sin\dfrac{x}{2} = \dfrac{ 3 }{ \sqrt{10} } } \\ &\sf{cos\dfrac{x}{2} = - \dfrac{1}{ \sqrt{10} } } \\ &\sf{tan\dfrac{x}{2} = - 3}  \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

Additional Information :- Trigonometry Formulas

sin(−θ) = −sin θ

cos(−θ) = cos θ

tan(−θ) = −tan θ

cosec(−θ) = −cosecθ

sec(−θ) = sec θ

cot(−θ) = −cot θ

Product to Sum Formulas

sin x sin y = 1/2 [cos(x–y) − cos(x+y)]

cos x cos y = 1/2[cos(x–y) + cos(x+y)]

sin x cos y = 1/2[sin(x+y) + sin(x−y)]

cos x sin y = 1/2[sin(x+y) – sin(x−y)]

Sum to Product Formulas

sin x + sin y = 2 sin [(x+y)/2] cos [(x-y)/2]

sin x – sin y = 2 cos [(x+y)/2] sin [(x-y)/2]

cos x + cos y = 2 cos [(x+y)/2] cos [(x-y)/2]

cos x – cos y = -2 sin [(x+y)/2] sin [(x-y)/2]

Sum or Difference of angles

cos (A + B) = cos A cos B – sin A sin B

cos (A – B) = cos A cos B + sin A sin B

sin (A+B) = sin A cos B + cos A sin B

sin (A -B) = sin A cos B – cos A sin B

tan(A+B) = [(tan A + tan B)/(1 – tan A tan B)]

tan(A-B) = [(tan A – tan B)/(1 + tan A tan B)]

cot(A+B) = [(cot A cot B − 1)/(cot B + cot A)]

cot(A-B) = [(cot A cot B + 1)/(cot B – cot A)]

cos(A+B) cos(A–B)=cos^2A–sin^2B=cos^2B–sin^2A

sin(A+B) sin(A–B) = sin^2A–sin^2B=cos^2B–cos^2A

Multiple and Submultiple angles

sin2A = 2sinA cosA = [2tan A /(1+tan²A)]

cos2A = cos²A–sin²A = 1–2sin²A = 2cos²A–1= [(1-tan²A)/(1+tan²A)]

tan 2A = (2 tan A)/(1-tan²A)