Question

prove that tan Alfa = sin 2 Alfa/1+cos2alfa and deduce the values of tan 15 degree and tan 22.5 degree.​

Answer 1

$\large\underline\blue{\bold{Given \: Question :- }}$

• Prove that tan Alfa = sin 2 Alfa/(1+cos2alfa) and deduce the values of tan 15 degree and tan 22.5 degree.

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:\large \blue{AηsωeR :} ✍$

$\large\underline\blue{\bold{Formula \: used:- }}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:(1). \: sin2x = 2sinx \: cosx$

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

$\bf \:(3). \: tanx = \dfrac{sinx}{cosx}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large\underline\blue{\bold{Understanding \: the \: Concept :- }}$

Here the concept of Trigonometric Identities is used. In this question mainly we will use 3 identities. Firstly using first identity and second identity, we shall simplify the equation to a easier form. Then using these, we will find the the main equation to be solved. We will keep on simplifying and then finally using third identity, we will get our answer.

Put the values of x to get the remaining values.

─━─━─━─━─━─━─━─━─━─━─━─━─

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

$\bf \:Let \: alpha = \: 'x'$

$\begin{gathered}\begin{gathered}\bf To \: find = \begin{cases} &\sf{Prove \: that \: tanx = \dfrac{sin2x}{1 + cos2x} } \\ &\sf{tan \: 15°} \\ &\sf{tan \: 22.5°} \end{cases}\end{gathered}\end{gathered}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:\large \red{AηsωeR : 1.} ✍$

$\bf \:\dfrac{sin2x}{1 + cos2x}$

$\bf \:  = \dfrac{2sinx \: cosx}{2 {cos}^{2}x }$

$\bf \: = \dfrac{sinx}{cosx}$

$\bf \: = tanx$

$\large{\boxed{\boxed{\bf{Hence, \: tanx = \dfrac{sin2x}{1 + cos2x}}}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:\large \red{AηsωeR : 2} ✍$

☆From first part, we have

$\bf \:  ⟼ tanx = \dfrac{sin2x}{1 + cos2x}$

$\bf \:Put \: x = 15°, \: we \: get$

$\sf \: ⟼tan15° = \dfrac{sin30°}{1 + cos30°}$

$\sf \: ⟼tan15° = \dfrac{\dfrac{1}{2} }{1 + \dfrac{ \sqrt{3} }{2} }$

$\sf \: ⟼tan15° =\dfrac{\dfrac{1}{2} }{\dfrac{2 + \sqrt{3} }{2} }$

$\sf \: ⟼tan15° =\dfrac{1}{2 + \sqrt{3} }$

☆On rationalize the denominator, we get

$\sf \: ⟼tan15° =\dfrac{1}{2 + \sqrt{3} } \times \dfrac{2 - \sqrt{3} }{2 - \sqrt{3} }$

$\sf \: ⟼tan15° =\dfrac{2 - \sqrt{3} }{ {2}^{2} - { (\sqrt{3} )}^{2} }$

$\sf \: ⟼tan15° =\dfrac{2 - \sqrt{3} }{4 - 3} = 2 - \sqrt{3}$

$\large{\boxed{\boxed{\bf{\bf\implies \:tan15° =2 - \sqrt{3} }}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\bf \:\large \red{AηsωeR : 3} ✍$

☆ From first part, we have

$\bf \:  ⟼ tanx = \dfrac{sin2x}{1 + cos2x}$

$\bf \:Put \: x = 22.5°, \: we \: get$

$\sf \: ⟼tan22.5° = \: \dfrac{sin45°}{1 + cos45°}$

$\sf \: ⟼tan22.5° =\dfrac{\dfrac{1}{ \sqrt{2} } }{1 + \dfrac{1}{ \sqrt{2} } }$

$\sf \: ⟼tan22.5° =\dfrac{\dfrac{1}{ \sqrt{2} } }{\dfrac{ \sqrt{2} + 1}{ \sqrt{2} } }$

$\sf \: ⟼tan22.5° =\dfrac{1}{ \sqrt{2} + 1 }$

☆On rationalize the denominator, we get

$\sf \: ⟼tan22.5° =\dfrac{1}{ \sqrt{2} + 1} \times \dfrac{ \sqrt{2} - 1}{ \sqrt{2} - 1}$

$\sf \: ⟼tan22.5° =\dfrac{ \sqrt{2} - 1 }{ {( \sqrt{2} )}^{2} - {1}^{2} }$

$\sf \: ⟼tan22.5° =\dfrac{ \sqrt{2} - 1}{1}$

$\sf \: ⟼tan22.5° = \sqrt{2} - 1$

$\large{\boxed{\boxed{\bf{Hence, \: tan22.5° = \sqrt{2} - 1}}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\large \green{\bf \: ⟼ Explore \: more } ✍$

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)

─━─━─━─━─━─━─━─━─━─━─━─━─