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

If $\sin { x } +\sin ^{ 2 }{ x } =1$, then the value of $\cos ^{ 12 }{ x } +3\cos ^{ 10 }{ x } +3\cos ^{ 8 }{ x } +\cos ^{ 6 }{ x } -2$ is equal to

  1. $0$
  2. $-1$
  3. $-2$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$sinx+sin^2x=1$

or, $sinx=1−sin^2x=cos^2x $

or, $sin^2x=cos^4x$

$cos^{12}x+3cos^{10}x+3cos^8x+cos^6x-2$

$=cos^6x(cos^6x+3cos^4x+3cos^2x+1)-2$

$=cos^6x((cos^2x)^3+3(cos^2x)^2.1+3.cos^2x.1^2+1^3)-2$

$=(cos^2x)^3(cos^2x+1)^3-2$

$=(cos^4x+cos^2x)^3-2$

$=(sin^2x+cos^2x)^3-2$

$=1^3-2$

$=1-2=-1$
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 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

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

  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

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

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

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

If $11 \sin^2 x + 7\cos^2x = 8$ then $x =$______

  1. $nx \pm \dfrac{\pi}{6},\forall n \in Z$
  2. $nx \pm \dfrac{\pi}{4},\forall n \in Z$
  3. $nx \pm \dfrac{\pi}{3},\forall n \in Z$
  4. $nx \pm \dfrac{\pi}{2},\forall n \in Z$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given $11\sin^2 x+7\cos^2 x=8$

$\implies 11\sin^2 x+7-7\sin^2 x=8$
$\implies 4\sin^2 x=1$
$\implies \sin^2 x=\dfrac{1}{4}$
$\implies \sin^2 x=\sin^2 \dfrac{\pi}{6}$
$\implies x=n\pi\pm \dfrac{\pi}{6},\forall Z$

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

If $\alpha, \beta$ are solution of equation a $cos \theta + b sin\theta = c$ then

  1. $sin \alpha + sin \beta = \dfrac{a^2-c^2}{b^2-a^2}$
  2. $cos \alpha + cos \beta = \dfrac{2ac}{a^2 + b^2}$
  3. $cos \alpha . cos \beta = \dfrac{c^2-b^2}{a^2 + b^2}$
  4. $\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$a\cos\theta+b\sin\theta=c$

$a^{2}\cos^{2}\theta+b^{2}\sin^{2}\theta+2ab\sin^{2}\theta.\cos\theta=c^{2}$

$a^{2}(1-\sin^{2}\theta)+b^{2}\sin^{2}\theta+2ab\sin\theta.\cos\theta=c^{2}$

$a^{2}-a^{2}\sin^{2}\theta+b^{2}\sin^{2}\theta\sin^{2}\theta+2ab\sin\theta\cos\theta=c^{2}$

$\sin^{2}\theta(b^{2}-a^{2})+2ab\sin\theta.\cos\theta+a^{2}-c^{2}=0$

So
$\sin \alpha.\sin \beta=\dfrac{a^{2}-c^{2}}{b^{2}-a^{2}}$
Multiple choice mathematics and statistics angle and their measurement angles and sides naming the sides in a right angled triangle understanding ratios

If $\cos x + cosy + \cos \theta = 0$ and $\sin x + \sin y + \sin \theta = 0$, then $\cot\left(\dfrac{x + y}{2}\right)$ 

  1. $\sin \theta$
  2. $\cos \theta$
  3. $\cot \theta$
  4. $\sin\left(\dfrac{x + y}{2}\right)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\cos x+\cos y+\cos \theta=0$

$\cos x+\cos y=-\cos \theta$
$\cfrac { 2\cos(x+y) }{ 2 } \times \cfrac { \cos(x-y) }{ 2 } =-\cos\theta  \longrightarrow 1$
$\sin x+\sin y+\sin \theta=0$
$\sin x+\sin y=-\sin \theta$
$\cfrac { 2\sin(x+y) }{ 2 } \times \cfrac { \sin(x-y) }{ 2 } =-\sin\theta $
Dividing both,
$\cfrac { \cfrac { 2\cos  (x+y) }{ 2 } \times \cfrac { \cos  (x-y) }{ 2 }  }{ \cfrac { 2\sin  (x+y) }{ 2 } \times \cfrac { \sin  (x-y) }{ 2 }  } =\cfrac { -\cos  \theta  }{ -\sin\theta } $
$\cfrac { \cot(x+y) }{ 2 } =\cot\theta $

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

If $sin:\theta +cos:\theta =p$ and $:tan:\theta +cot:\theta =q$ then $:q\left(p^2-1\right)=$

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

$sin:\theta +cos:\theta =p$ 
Squaring on both side we get
$\Rightarrow 1+sin:2\theta =p^2$    $[\because sin:2\theta =p^2-1]$
$Tan\theta +\frac{1}{Tan\theta :}=q$

$\frac{Tan^2\theta +1}{2Tan\theta :}=\frac{q}{2}$

$cosec:2\theta =\frac{q}{2}$

$\Rightarrow \frac{1}{p^2-1}=\frac{q}{2}$   $\Rightarrow 2=\left(p^2-1\right)q$

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

If $\tan { \theta  } .\tan { (120-\theta ) } .\tan { (120+\theta ) } =\dfrac { 1 }{ \sqrt { 3 }  }$, then $\theta $

  1. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 18 } ,n\epsilon Z$
  2. $\dfrac { n\pi }{ 3 } +\cfrac { \pi }{ 12 } ,n\epsilon Z$
  3. $\dfrac { n\pi }{ 12 } +\dfrac { \pi }{ 12 } ,n\epsilon Z$
  4. $\dfrac { n\pi }{ 3 } +\dfrac { \pi }{ 6 } ,n\epsilon Z$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the identity tan(theta) * tan(60-theta) * tan(60+theta) = tan(3*theta), we have tan(3*theta) = 1/sqrt(3). Thus, 3*theta = n*pi + pi/6, which simplifies to theta = n*pi/3 + pi/18.

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

The value of sin $15^0$ is

  1. $\dfrac{\sqrt{3}+1}{2}$
  2. $\dfrac{\sqrt{3}+1}{2\sqrt{2}}$
  3. $\dfrac{-(\sqrt{3}+1)}{2\sqrt{2}}$
  4. $\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\sin15^o$


$=\sin(45^o-30^o)$

$=\sin45^o \ \cos30^o - \cos45^o \ \sin30^o$

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

$=\dfrac{\sqrt{3}-1}{2\sqrt{2}}$
Hence answer is D

Multiple choice the nth roots of unity complex numbers maths

The value of $\sum _{ n=1 }^{ 10 }{ \left( sin\frac { 2n\pi  }{ 11 } -icos\frac { 2n\pi  }{ 11 }  \right)  } $

  1. $i$
  2. $-i$
  3. $0$

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

The sum is a geometric series of complex numbers. Using the identity for the sum of roots of unity, the sum evaluates to i.