Mathematics

Trigonometric Identities and Equations

223 Questions

Solve a variety of questions based on trigonometric identities and mathematical equations. Key areas covered include double angle formulas, the law of sines, and quadrant angles. This material helps students preparing for advanced mathematics exams, UPSC, and state public service commissions.

Double angle formulasLaw of sinesTrigonometric quadrantsLaplace transformSecant functionTaylor series

Trigonometric Identities and Equations Questions

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

For the function $\sin\pi x$ centred at $a=0.5$.using taylor series expansion,find approximate value of $\sin\left(\dfrac{\pi}{2} + \dfrac{\pi}{10} \right)$

  1. $0.9511$
  2. $0.9633$
  3. $0.8962$
  4. $0.2134$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The taylor series is given by $\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$

$f(x)\approx\sum _{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k=\sum _{k=0}^{2}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$
Finally after simplifying, we get
$f(x)\approx\dfrac{1}{0!}\left (x-\dfrac{1}{2}\right)^0+\dfrac{0}{1!}\left (x-\dfrac{1}{2}\right)^1+\dfrac{-(\pi)^2}{2!}(x-\dfrac{1}{2})^2$ 
$f(x)=1-\dfrac{(\pi)^2}{2}(x-\dfrac{1}{2})^2$
$\sin(\pi x)=1-\dfrac{\pi^2}{2}\left (x-\dfrac{1}{2}\right)^2$
$\sin\left (\pi \left (\dfrac{1}{2}+\dfrac{1}{10}\right)\right)=1-\dfrac{\pi^2}{2}\left (\dfrac{1}{2}+\dfrac{1}{10}-\dfrac{1}{2}\right)^2=0.95065\approx0.951$

Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $\sin { { y+e }^{ -x\cos { y }  } } =e\quad then\quad \frac { dy }{ dx } \quad at\quad (1,\pi )$ is equal to 

  1. $\sin { y } $
  2. $-x\cos { y } $
  3. $e$
  4. $\sin { y } -x\cos { y } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

$\sin y+e^{-x\cos y}=e$

$\cos y \dfrac{dy}{dx}+e^{-x\cos y}\left [ x\sin y \dfrac{dy}{dx}-\cos y \right ]=0$

$\Rightarrow \cos y \dfrac{dy}{dx}+e^{-x\cos y} x\sin y \dfrac{dy}{dx} -\cos ye^{-x\cos y}=0$

$\dfrac{dy}{dx}[\cos y+x\sin y e^{-x\cos y}]=\cos y e^{-x\cos y}$

$\dfrac{dy}{dx}=\dfrac{\cos y e^{-x\cos y}}{\cos y+x\sin y e^{-x\cos y}}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{\cos \pi  e^{-\cos \pi}}{\cos \pi+1 \sin \pi e^{-\cos \pi }}$

$\left [ \dfrac{dy}{dx} \right ] _{(1, \pi )}=\dfrac{-1 \times e}{-1+0 \times e}=e$
Multiple choice maths differencial calculus - differenciability and methods of differnciation differentiation by substitution methods of differentiation derivative of a function

If $x=a\sin \theta$ and $y=b\cos\theta$, then $\displaystyle\frac{d^2y}{dx^2}$ is 

  1. $\displaystyle\frac{a}{b^2}\sec^2\theta$
  2. $\displaystyle\frac{-b}{a}\sec^2\theta$
  3. $\displaystyle\frac{b}{a^2}\sec^3\theta$
  4. $\displaystyle\frac{-b}{a^2}\sec^3\theta$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given, $x=a\sin\theta$ and $y=b\cos\theta$
On differentiating w.r.t.$\theta$, we get
$\displaystyle\frac{dx}{d\theta}=a\cos\theta$
and $\displaystyle\frac{dy}{d\theta}=-b\sin \theta$
$\Rightarrow \displaystyle\frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=-\frac{b}{a}\tan\theta$
Again, differentiating w.r.t. $x$, we get
$\displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{d\theta}{dx}$
$\Rightarrow \displaystyle\frac{d^2y}{dx^2}=-\frac{b}{a}sec^2\theta\cdot\frac{1}{a\cos\theta}$
$=-\displaystyle\frac{b}{a^2}sec^3\theta$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If $A+B=\dfrac{\pi}{3}$ and $\cos{A}+\cos{B}=1$, then which of the following is true

  1. $\cos{\left(A-B\right)}=\dfrac{1}{3}$
  2. $\left|\cos{A}-\cos{B}\right|=\sqrt{\dfrac{2}{3}}$
  3. $\cos{\left(A-B\right)}=-\dfrac{1}{3}$
  4. $\left|\cos{A}-\cos{B}\right|=\dfrac{1}{2\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$\cos{A}+\cos{B}=1$


$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}=1$

Since $A+B=\dfrac{\pi}{3}\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{6}$
Hence $\cos{\left(\dfrac{A+B}{3}\right)}=\cos{\left(\dfrac{\pi}{6}\right)}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow 2\cos{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{\dfrac{\sqrt{3}}{2}}$

$\Rightarrow \cos{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{\sqrt{3}}$

Squaring both sides, we get

${\cos}^{2}{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{3}$

$\Rightarrow 2{\cos}^{2}{\left(\dfrac{A-B}{2}\right)}=\dfrac{2}{3}$

$\Rightarrow 2{\cos}^{2}{\left(\dfrac{A-B}{2}\right)}-1=\dfrac{2}{3}-1=\cos {(A-B)}=\dfrac{-1}{3}$

$\left|\cos{A}-\cos{B}\right|=2\sin{\left(\dfrac{A+B}{2}\right)}\sin{\left(\dfrac{B-A}{2}\right)}$

                    $=2\times\dfrac{1}{2}\sqrt{1-\dfrac{1}{3}}$

                    $=\sqrt{\dfrac{2}{3}}$ (on simplification)

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If  $\alpha$  is the angle which each side of a regular polygon of  $n$  sides subtends at its centre, then  $1 + \cos \alpha + \cos 2 \alpha + \cos 3 \alpha \ldots + \cos ( n - 1 ) \alpha$  is equal to

  1. $n$
  2. $0$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a regular polygon of n sides with central angle alpha = 2pi/n, the sum 1 + cos(alpha) + cos(2alpha) + ... + cos((n-1)alpha) represents the real part of the sum of roots of unity, which geometrically sums to zero because the vectors representing the sides form a closed polygon.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The sum of the radii of inscribed and circumscribed circles of an n sided regular polygon of side $'a'$ is

  1. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/2x} + \cot \frac{\pi}{x} \right )$
  2. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{2x} \right )$
  3. $=\frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$R\sin \theta  = \frac{a}{2}$
$R = \frac{a}{2\sin \theta }$
$\tan \theta = \frac{a/2}{r}$
$r = \frac{a}{2\tan \theta }                                   \theta = \frac{2\pi}{n} \times\frac{1}{2}$
$R+r = \frac{a}{2} \left ( \frac{1}{\sin \theta}+\frac{\sin \theta}{\cos \theta } \right )             = \frac{\pi}{x}$
    $= \frac{a}{2} \left ( \frac{1}{\sin \pi/x} + \cot \frac{\pi}{x} \right )$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If ${A} _{1}{A} _{2}{A} _{3}...{A} _{n}$ be a regular polygon of $n$ sides and 
$\dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}},$then

  1. $n=5$
  2. $n=6$
  3. $n=7$

  4. none of these.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If radius of circle is $r$ then 
${A} _{1}{A} _{2}=2r\sin{\left(\dfrac{\pi}{n}\right)}$
${A} _{1}{A} _{3}=2r\sin{\left(\dfrac{2\pi}{n}\right)}$
${A} _{1}{A} _{4}=2r\sin{\left(\dfrac{3\pi}{n}\right)}$
$\because \dfrac{1}{{A} _{1}{A} _{2}}=\dfrac{1}{{A} _{1}{A} _{3}}+\dfrac{1}{{A} _{1}{A} _{4}}$
$\Rightarrow \dfrac{1}{2r\sin{\left(\dfrac{\pi}{n}\right)}}=\dfrac{1}{2r\sin{\left(\dfrac{2\pi}{n}\right)}}+\dfrac{1}{2r\sin{\left(\dfrac{3\pi}{n}\right)}}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{3\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}+\sin{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}\left[\sin{\left(\dfrac{3\pi}{n}\right)}-\sin{\left(\dfrac{\pi}{n}\right)}\right]=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
Using transformation angle formula, we get
$\Rightarrow \sin{\left(\dfrac{2\pi}{n}\right)}.2\cos{\left(\dfrac{2\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}\sin{\left(\dfrac{\pi}{n}\right)}$
$\Rightarrow 2\sin{\left(\dfrac{2\pi}{n}\right)}\cos{\left(\dfrac{2\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
Using multiple angle formula, $2\sin{A}\cos{A}=\sin{2A}$ we get
$\sin{\left(\dfrac{4\pi}{n}\right)}=\sin{\left(\dfrac{3\pi}{n}\right)}$
$\therefore \dfrac{4\pi}{n}=r+{\left(-1\right)}^{r}\dfrac{3}{n}$ for $r=1,n=7$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The sum of inradius and circumradius of incircle and circumcircle of a regular polygon of side $n$ is

  1. $\dfrac {a}{4}\cot \dfrac {\pi}{2n}$
  2. $a\cot \dfrac {\pi}{n}$
  3. $\dfrac {a}{2} \cot \dfrac {\pi}{2n}$
  4. $a\cot \dfrac {\pi}{2n}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$r + R = \dfrac {a}{2}\cot \dfrac {\pi}{n} + \dfrac {a}{2}cosec \dfrac {\pi}{n}$
$= \dfrac {a}{2} \left (\dfrac {1 + \cos \frac{\pi}{n}}{\sin \frac{\pi}{n}}\right ) = \dfrac {a}{2} \dfrac {2\cos^{2} \dfrac {\pi}{2n}}{2\sin \dfrac {\pi}{2n}\cdot \cos \dfrac {\pi}{2n}}$
$= \dfrac {a}{2} \cot \dfrac {\pi}{2n}$.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The sum of the radii of inscribed and circumscribed circles of an $n$ -sided regular polygon with side equal to one unit is?

  1. $\displaystyle \frac{1}{2}\cot \frac{\pi }{2n}$
  2. $\displaystyle \cot \frac{\pi }{2n}$
  3. $\displaystyle \cot \frac{\pi }{n}$
  4. $\displaystyle \frac{1}{2}\tan \frac{\pi }{2n}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

From the figure:
Side of polygon $(AB)=1$
$AO=\dfrac { 1 }{ 2 } $
$\angle O=\dfrac { \pi  }{ 2n } $

In right angled $\triangle COA$ :
$\sin { O } =\dfrac { AC }{ AO } $
$\Rightarrow \sin { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2R } $       ..(1)

$\tan { O } =\dfrac { AC }{ CO } $
$\Rightarrow \tan { \dfrac { \pi  }{ n }  } =\dfrac { 1 }{ 2r } $       ...(2)

From (1) and (2)
$R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ \sin { \dfrac { \pi  }{ n }  }  } +\dfrac { 1 }{ \tan { \dfrac { \pi  }{ n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \left( \dfrac { 1+\cos { \dfrac { \pi  }{ n }  }  }{ \sin { \dfrac { \pi  }{ n }  }  }  \right) =\dfrac { 1 }{ 2 } \left( \dfrac { 2\cos ^{ 2 }{ \dfrac { \pi  }{ 2n }  }  }{ 2\cos { \dfrac { \pi  }{ 2n }  } \sin { \dfrac { \pi  }{ 2n }  }  }  \right) $

$\Rightarrow R+r=\dfrac { 1 }{ 2 } \cot { \dfrac { \pi  }{ 2n }  } $

Ans: A

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

If $\sin \theta + \cos \theta = 1$, then what is the value of $\sin \theta \cos \theta$?

  1. $2$
  2. $0$
  3. $1$
  4. $\dfrac {1}{2}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given, $\sin \theta + \cos \theta = 1$

Squaring both sides gives,
$\sin^{2}\theta + \cos^{2}\theta + 2\sin \theta \cos \theta = 1$
$\Rightarrow 1 + 2\sin \theta \cos \theta = 1$
$\Rightarrow 2\sin \theta \cos \theta = 0$
$\Rightarrow \sin \theta \cos \theta = 0$

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

For a
positive integer n,
let
${f _n}\left( \theta  \right) = \left( {\tan \frac{\theta }{2}} \right)\left( {1 + \sec \theta } \right)\left( {1 + \sec 2\theta } \right)\left( {1 + \sec {2^2}\theta } \right)...\left( {1 + \sec {2^n}\theta } \right),then$

  1. ${f _2}\left( {\frac{\pi }{{16}}} \right) = 1$
  2. ${f _3}\left( {\frac{\pi }{{32}}} \right) = 1$
  3. ${f _4}\left( {\frac{\pi }{{64}}} \right) = 1$
  4. ${f _5}\left( {\frac{\pi }{{128}}} \right) = 1$
Reveal answer Fill a bubble to check yourself
A,B,C,D Correct answer
Explanation

$f _n(\theta)=(\tan \frac{\theta}{2})(1+\text{sec}\theta)(1+\text{sec} 2\theta)(1+\text{sec}2^2 \theta)\cdots(1+\text{sec}2^n \theta)$

           $=\dfrac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\times \dfrac{1+\cos \theta}{\cos \theta}\times \dfrac{1+\cos 2\theta}{\cos 2\theta}\times \dfrac{1+\cos 2^2 \theta}{\cos 2^2 \theta}\cdots\times \dfrac{1+\cos 2^n \theta}{\cos 2^n \theta}$
           $=\dfrac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}\times \dfrac{2\cos ^2 \frac{\theta}{2}}{\cos \theta}\times \dfrac{2\cos ^2 \theta}{\cos 2 \theta}\times \dfrac{2\cos^2 2\theta}{\cos 2^2 \theta}\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos^2 2^n\theta}$
           $=(2\sin \frac{\theta}{2}\cos \frac{\theta}{2})\times (2\cos \theta)\times (2\cos 2 \theta)\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos 2^n \theta}$
           $=(2\sin \theta\cos \theta)\times (2\cos 2 \theta)\times \cdots\times \dfrac{2\cos 2^{n-1}\theta}{\cos 2^n \theta}$
           $=\dfrac{\sin 2^n \theta}{\cos 2^n \theta}=\tan 2^n \theta$

$f _2\bigg(\dfrac{\pi}{16}\bigg)=\tan (2^2\times \dfrac{\pi}{16})=\tan \frac{\pi}{4}=1$
$f _3\bigg(\dfrac{\pi }{32}\bigg)=\tan (2^3\times \dfrac{\pi}{32})=\tan \frac{\pi}{4}=1$
$f _4\bigg(\dfrac{\pi}{64}\bigg)=\tan (2^4\times \dfrac{\pi}{64})=\tan \frac{\pi}{4}=1$
$f _5\bigg(\dfrac{\pi}{128}\bigg)=\tan (2^5\times \dfrac{\pi}{128})=\tan \frac{\pi}{4}=1$

Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

$8\sin { \theta  } \cos { \theta  } .\cos { 2\theta  } \cos { 4\theta  } =\sin { x } \Longrightarrow x=$?

  1. <span class="MathJax_Preview"><span class="MathJax"><span class="math"><span class="mrow"><span class="mi">x<span class="mo">=-<span class="mn">8<span class="mi">θ<span class="MJX_Assistive_MathML">x=8θ
  2. $x=8\theta$

  3. <span class="MathJax_Preview"><span class="MathJax"><span class="math"><span class="mrow"><span class="mi">x<span class="mo">=4<span class="mi">θ<span class="MJX_Assistive_MathML">x=8θ

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$8\sin\theta \cos\theta\cos 2\theta \cos 4\theta =\sin x$

$=4\sin 2 \theta\cos 2\theta\cos 4 \theta$
$=2\sin 4\theta\cos 4\theta$
$\sin 8\theta=x$
$x=8\theta$