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 using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

If $\tan \theta =\dfrac {\cos 9^{o}+\sin 9^{o}}{\cos 9^{o}-\sin 9^{o}}$, then the value of $\theta$ is

  1. $9^{o}$
  2. $54^{o}$
  3. $18^{o}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Divide numerator and denominator by cos(9 degrees) to get tan(theta) = (1 + tan(9 degrees)) / (1 - tan(9 degrees)) = tan(45 degrees + 9 degrees) = tan(54 degrees). Thus, theta = 54 degrees.

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

The value of expression $\dfrac { 2\left( \sin{ 1 }^{ o }+\sin{ 2 }^{ o }+\sin{ 3 }^{ o }+.....+\sin{ 89 }^{ o } \right)  }{ 2\left( \cos{ 1 }^{ o }+\cos{ 2 }^{ o}+......+\cos{ 44 }^{ o } \right) +1 }$ equals

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

The numerator is a sum of sines from 1 to 89, which equals cot(1/2) * sin^2(45). The denominator simplifies similarly. The ratio evaluates to sqrt(2).

Multiple choice using trigonometric tables trigonometric ratios of some specific angles trigonometric identities trigonometry maths

If $sin(A-B)=\frac { 1 }{ 2 } ,cos(A+B)=\frac { 1 }{ 2 } ,{ 0 }^{ 0 }<A+B\le { 90 }^{ 0 }$ then A =

  1. $15^{ 0 }$
  2. $45^{ 0 }$
  3. $90^{ 0 }$
  4. $30^{ 0 }$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have sin(A - B) = 1/2, so A - B = 30 degrees. We also have cos(A + B) = 1/2, so A + B = 60 degrees. Adding these two equations gives 2A = 90 degrees, which means A = 45 degrees.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points $a(cos \alpha + i sin \alpha)$ , $b(cos \beta + i sin \beta)$ and $c(cos \gamma + isin \gamma)$ are collinear then the value of $|z|$ is:  
( where ${z = bc  \ sin(\beta-\gamma) + ca \ sin(\gamma-\alpha) + ab \ sin(\alpha - \beta) + 3i -4k}$ )

  1. $2$
  2. $5$
  3. $1$

  4. None of these.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given $a=cos\alpha+i\sin\alpha=e^{i\alpha}$ , $b=cos\beta+isin\beta=e^{i\beta}$ and $c=cos\gamma+isin\gamma=e^{i\gamma}$
Consider $bcsin(\beta-\gamma)=e^{i(\beta+\gamma)}sin(\beta-\gamma)=\frac{1}{2i}e^{i(\beta+\gamma)}(e^{i(\beta-\gamma)}-e^{-i(\beta-\gamma)}) = \frac{1}{2i}(e^{i(2\beta)}-e^{i(2\gamma)})$
Similarly we get $casin(\gamma-\alpha) = \frac{1}{2i}(e^{i(2\gamma)}-e^{i(2\alpha)})$ and $absin(\alpha-\beta) = \frac{1}{2i}(e^{i(2\alpha)}-e^{i(2\beta)})$
Therefore we get $bcsin(\beta-\gamma)+casin(\gamma-\alpha)+absin(\alpha-\beta)=0$
So we get $z=3i-4i$
$\Rightarrow |z|=5$
Multiple choice maths squares and square roots finding the square of a number finding square of a number patterns in square numbers

If sin$\theta -cosec  \theta =\sqrt{5},$ then the value of sin  $\theta  + cosec  \theta$ is:

  1. $\sqrt{3}$

  2. 1

  3. 3

  4. 9
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\Rightarrow \sin\theta-cosec\theta=\sqrt{5}$


$\Rightarrow \sin\theta-\dfrac{1}{\sin\theta}=\sqrt{5}$      $(\because cosec\theta=\dfrac{1}{\sin\theta})$

$\Rightarrow \sin^2\theta-\sqrt{5}\sin\theta-1=0$

Solving equation to get roots.

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{\pm3+\sqrt{5}}{2}$ (substitute values to get roots)

To find:-

$\sin\theta+cosec\theta$

Using $\dfrac{3+\sqrt{5}}{2}$

$=\dfrac{3+\sqrt{5}}{2}+\dfrac{2}{3+\sqrt{5}}$

$=\dfrac{(3+\sqrt{5})^2+4}{2(3+\sqrt{5})}$

$=\dfrac{9+4+5+6\sqrt{5}}{2(3+\sqrt{5})}$

$=\dfrac{6(3+\sqrt{5})}{2(3+\sqrt{5})}$

$=3$


Using $\dfrac{-3+\sqrt{5}}{2}$

$=\dfrac{-3+\sqrt{5}}{2}+\dfrac{2}{-3+\sqrt{5}}$

$=\dfrac{(-3+\sqrt{5})^2+4}{2(-3+\sqrt{5})}$

$=\dfrac{9+4+5-6\sqrt{5}}{2(-3+\sqrt{5})}$

$=\dfrac{-6(-3+\sqrt{5})}{2(-3+\sqrt{5})}$

$=-3$


According to option answer is $3$

Multiple choice business maths applications of matrices and determinants non-homogeneous linear equations system of simultaneous equations matrices

The values of $\theta $ lying between $\theta =0$ and $\theta =\dfrac {\pi}{2}$ and satisfying the equation
$\begin{vmatrix}
1+\sin ^{2}\theta  & \cos ^{2}\theta  & 4\sin 6\theta \
\sin ^{2}\theta  & 1+\cos ^{2}\theta  & 4\sin 6\theta \
\sin ^{2}\theta  & \cos ^{2}\theta  & 1+4\sin 6\theta
\end{vmatrix}$
are given by

  1. $\dfrac {\pi }{36}, \dfrac{5\pi}{ 36}$
  2. $\dfrac{7\pi}{36}, \dfrac{11\pi}{3}$
  3. $\dfrac{5\pi }{36}, \dfrac{7\pi }{36}$
  4. $\dfrac{11\pi}{36}, \dfrac{\pi }{36}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{vmatrix} 1+\sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & 1+\cos ^{ 2 } \theta  & 4\sin  6\theta  \ \sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 1+4\sin  6\theta  \end{vmatrix}=0$

Applying ${ R } _{ 3 }\rightarrow { R } _{ 3 }-{ R } _{ 1 },{ R } _{ 2 }\rightarrow { R } _{ 2 }-{ R } _{ 1 }$

$\Rightarrow \begin{vmatrix} 1+\sin ^{ 2 } \theta  & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ -1 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix}=0$

Applying ${ C } _{ 1 }\rightarrow { C } _{ 1 }+{ C } _{ 2 }$

$\Rightarrow \begin{vmatrix} 2 & \cos ^{ 2 } \theta  & 4\sin  6\theta  \ 0 & 1 & 0 \ -1 & 0 & 1 \end{vmatrix}=0\ \Rightarrow 2+4\sin  6\theta =0\Rightarrow \sin  6\theta =-\cfrac { 1 }{ 2 } \ \Rightarrow 6\theta =n\pi +{ \left( -1 \right)  }^{ n }\left( -\cfrac { \pi  }{ 6 }  \right) \ \Rightarrow \theta =\cfrac { n\pi  }{ 6 } +{ \left( -1 \right)  }^{ n+1 }\left( \cfrac { \pi  }{ 36 }  \right) \ \Rightarrow \theta =\cfrac { 7\pi  }{ 36 } ,\cfrac { 11\pi  }{ 36 } $

Multiple choice maths algebraic functions, equations and inequalities descartes rule sign of quadratic expression sum and product of the roots of a polynomial equation

If $o<\alpha<\beta<\gamma<\dfrac {\pi}{2}$, then the equation $\dfrac {1}{x-\sin \alpha}+\dfrac {1}{x-\sin\beta}+\dfrac {1}{x-\sin \gamma}=0$ has

  1. Imaginary roots

  2. Real and equal roots

  3. Real and unequal roots

  4. Rational roots
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} \frac { 1 }{ { x-\sin  \alpha  } } +\frac { 1 }{ { x-\sin  \beta  } } +\frac { 1 }{ { x-\sin  \gamma  } } =0 \ 0<\alpha <\beta <\gamma <\frac { \pi  }{ 2 }  \ let\, \alpha ={ 30^{ 0 } },\beta ={ 45^{ 0 } },\gamma ={ 60^{ 0 } } \ \Rightarrow \frac { 1 }{ { x-\frac { 1 }{ 2 }  } } =\frac { 1 }{ { x-2\sqrt { 2 }  } } =\frac { 1 }{ { x-\sqrt { \frac { 3 }{ 2 }  }  } } =0 \end{array}$

hence roots are real and unequal

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $R \subset\left ( 0,\pi  \right )$ denote the set of values of which satisfies the equation $ \displaystyle 2^{\left ( 1+\left | \cos x \right |+\left | cos^{2}x \right |+\left | cos^{3}x \right | \right )+\left | cos^{4}x  \right |...............\infty}=4$ then $R$ equals

  1. $\displaystyle\left \{ -\frac{\pi }{3} \right \}$
  2. $\displaystyle\left \{ \frac{\pi }{3},\frac{2\pi }{3} \right \}$
  3. $\displaystyle\left \{ \frac{-\pi }{3},\frac{2\pi }{3} \right \}$
  4. $\displaystyle\left \{ \frac{\pi }{3},\frac{-2\pi }{3} \right \}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

${ 2 }^{ \left( 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty  \right)  }={ 2 }^{ 2 }\ \Rightarrow 1+\left| \cos { x }  \right| +\left| \cos ^{ 2 }{ x }  \right| +.........\infty =2\ \Rightarrow \dfrac { 1 }{ 1-\left| \cos { x }  \right|  } =2\ \Rightarrow 1-\left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow \left| \cos { x }  \right| =\dfrac { 1 }{ 2 } \ \Rightarrow x=\dfrac { \pi  }{ 3 } ,\dfrac { 2\pi  }{ 3 } $
  in the range $\left( 0,\pi  \right) $

Multiple choice business maths matrix properties of matrix multiplication properties of multiplication of matrix multiplication of matrices

If $A = \left[ \begin{array}{l}\cos \theta \,\,\,\,\sin \theta \ - \sin \theta \,\,\,\cos \theta \end{array} \right]$ where $\theta  = \frac{{2\pi }}{{19}}$ then ${A^{2017}} = $

  1. $A$
  2. ${A^3}$
  3. ${A^5}$
  4. $i$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A is a rotation matrix R(theta). A^n = R(n*theta). Here, A^2017 = R(2017 * 2pi / 19). Since 2017 = 19 * 106 + 3, A^2017 = R(3 * 2pi / 19) = A^3.

Multiple choice mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If $\int\dfrac{2\cos x-\sin x+\lambda}{\cos x-\sin x-2}dx=A In\left|\cos x+\sin x-2\right|+Bx+C$. Then the ordered triplet $\left(A,B,\lambda\right)$, is 

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

Perform the integration by expressing the numerator as a linear combination of the denominator and its derivative. Let 2cos(x) - sin(x) + lambda = A(cos(x) - sin(x) - 2) + B(-sin(x) - cos(x)). Solving for coefficients yields A=1/2, B=3/2, lambda=-1.

Multiple choice maths polygons exterior angles of polygon sum of exterior angles of polygons exterior angles of a polygon

If $B$ the exterior angle of a regular polygon of $n-sides$ and $A$ is any constant then $\cos A + \cos (A + B) + \cos (A + 2B) + .... n$ terms is equal to:

  1. $0$
  2. $\cos A$
  3. $1$
  4. $\dfrac {\sqrt {3}}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The sum of exterior angles of a polygon is $ 2\pi$ or $ 360^{\circ}$

If it is a regular polygon

Exterior Angles $ = \dfrac{2\pi}{n} $ 

n is no of sides

According to question

$ B = \dfrac{2\pi}{n}.$

$ \cos A + \cos (A+B) + \cos (A+2B)+ ...n+ terms $

$ = \cos A + \cos\left ( A + \dfrac{2 \pi}{n} \right )+ \cos \left ( A+2 \left ( \dfrac{2\pi}{n} \right ) \right )+...+ \cos \left ( A+ (n-2) \dfrac{2\pi}{n} \right ) + \cos \left ( A+(n-1) \dfrac{2\pi}{n} \right )$

$ \cos A+ \cos \left ( A+\dfrac{2\pi}{n} \right ) + \cos \left ( A+2 \left ( \dfrac{2\pi}{n} \right ) \right )+ ...+ \cos \left ( A+n \left ( \dfrac{2\pi}{n} \right )-2\left ( \dfrac{2\pi}{n} \right )\right ) + \cos \left ( A+n \left ( \dfrac{2\pi}{n} \right )-\dfrac{2\pi}{n}\right )$

$ = \cos A + \cos \left ( A +\dfrac{2\pi}{n} \right )+ \cos \left ( A+2 \left ( \dfrac{2\pi}{n} \right ) \right ) + ...+ \cos \left ( A+2\pi -2\left ( \dfrac{2\pi}{n} \right ) \right ) + \cos \left ( A + 2\pi- \dfrac{2\pi}{n} \right )$

$ = \cos A + \cos \left ( A+\dfrac{2\pi}{n} \right )+ \cos \left ( A+2\left ( \dfrac{2\pi}{n} \right ) \right ) + ...+ \cos\left ( A-2 \left ( \dfrac{2\pi}{n} \right ) \right )+ cos \left ( A - \dfrac{2\pi}{n} \right )$

$ \left \{ \because \cos (2\pi - \theta) = \cos \theta \right \}$

$ = \cos A + \cos (A) \, \cos \left ( \dfrac{2\pi}{n} \right )- \sin(A) . \sin\left ( \dfrac{2\pi}{n} \right )+ \cos A. \cos \dfrac{4\pi}{n}- \sin A $

$ \sin \dfrac{4\pi}{n}+ ...+ \cos A. \cos \dfrac{4\pi}{n}+ \sin A\, \sin \dfrac{4\pi}{n} + \cos A. \cos \dfrac{2\pi}{n} + \sin A . \sin \dfrac{\pi}{n}$

$ = \cos\,A + \cos\,A. \cos\dfrac{2\pi}{n}+ \cos A \,\cos\dfrac{4\pi}{n} + ... \cos A\, \cos \dfrac{4 \pi}{n} + \cos A.\cos \dfrac{2\pi}{n}.$

$ = \cos A \left \{ 1+ \cos\dfrac{2\pi}{n} + \cos \dfrac{4\pi}{n}+ ... \cos\dfrac{4\pi}{n}+ \cos\dfrac{2\pi}{n} \right \}$

$ = \cos A (1-1)$

$ =0 $