Mathematics · Quantitative Aptitude

Trigonometric Identities and Values

70 Questions

Trigonometric identities and values questions require evaluating complex angles using standard formulas. These are critical for quantitative aptitude sections in SSC and various state exams. Success depends on memorizing standard values and applying transformation rules.

angle transformationstrigonometric ratiosstandard angle valuessine cosine products

Trigonometric Identities and Values Questions

Multiple choice general knowledge

cos60=

  1. 1/2

  2. 0

  3. 1

  4. sqrt3/2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

cos(60°) = 1/2. In the special 30-60-90 right triangle, the side adjacent to the 60° angle is half the hypotenuse. cos(0°) = 1 and cos(90°) = 0, so 1/2 is the correct intermediate value.

Multiple choice general knowledge

value of sinx where x=53 degree

  1. .75

  2. .65

  3. .80

  4. .60
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

sin(53°) ≈ 0.7986, which rounds to 0.80. You can verify this using a calculator or trigonometric tables. The other values (0.75, 0.65, 0.60) do not match the actual sine of 53 degrees.

Multiple choice general knowledge

value of cos37

  1. .80

  2. .60

  3. .75

  4. .732
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

cos(37°) ≈ 0.7986, which rounds to 0.80. This is a standard trigonometric value that can be verified with a calculator. Option D (0.732) is actually closer to cos(43°).

Multiple choice maths trigonometry trigonometric ratios of acute angles compound angles, multiple angles, sub multiple angles and transformation formulae trigonometric identities

$\sin\ (45^{o}+\theta)-\cos\ (45^{o}-\theta)$ is equal to

  1. $2\cos \theta$
  2. $0$
  3. $2\sin \theta$
  4. $1$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the co-function identity, cos(90 - x) = sin x, we can rewrite cos(45 - theta) as sin(45 + theta). Therefore, sin(45 + theta) - cos(45 - theta) becomes sin(45 + theta) - sin(45 + theta) = 0.

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

The value of $\cot 15^{\circ} \cot 20^{\circ} \cot 70^{\circ} \cot 75^{\circ}$ is equal to

  1. $-1$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The value of $\cot 15^{\circ}.\cot 20^{\circ}.\cot 70^{\circ}.\cot 75^{\circ}$ is
$= (\cot 15^{\circ} . \cot 75^{\circ}) (\cot 20^{\circ} \cot 70^{\circ})$

Now $\cot 75^{\circ} = \tan 15^{\circ}$ and $\cot 70^{\circ} = \tan 20^{\circ}$   ....................... $\cot(90-\theta)=\tan\theta$

Therefore, the given expression can be written as 
$ (\cot 15^{\circ} \tan 15^{\circ})(\cot 20^{\circ} \tan 20^{\circ})$
$= 1\times 1 = 1$

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

The value of $cos^{2}30^{0}-cos^{2}60^{0}-cos 60^{0}$ is

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

$cos^{2}30^{0}-cos^{2}60^{0}-cos 60^{0}={ \left( \frac { \sqrt { 3 }  }{ 2 }  \right)  }^{ 2 }-{ \left( \frac { 1 }{ 2 }  \right)  }^{ 2 }-\frac { 1 }{ 2 } =\frac { 3 }{ 4 } -\frac { 1 }{ 4 } -\frac { 1 }{ 2 } =\frac { 3-1-2 }{ 4 } =0$

Multiple choice mathematics and statistics angle and their measurement degree measure of angle measure of angle radians or degrees

The value of $\dfrac{1}{\cos 290^o}+\dfrac{1}{\sqrt{3}\sin 250^o}$ is?

  1. $\dfrac{2\sqrt{3}}{3}$
  2. $\dfrac{4\sqrt{3}}{3}$
  3. $\sqrt{3}$

  4. None
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Here, $\dfrac{1}{ \cos 290^{o}} + \dfrac{1 }{ \sqrt{3} \sin 250^{o}}$ 
$= \dfrac{1}{ \cos (270+20)^{o}} + \dfrac{1}{ \sqrt{3} \sin (270-20)^{o}}$
as we know, $\cos (270+A)= \sin A$
& $\sin (270- B)= - \cos B$
So, $=\dfrac{1}{\sin 20}+ \dfrac{1}{\sqrt{3} (- \cos 20)}$
$=\dfrac{- \sqrt{3} \cos 20+ \sin 20}{- \sqrt{3} \cos 20 \cos 20}$
$=\dfrac{- (\sqrt{3} \cos 20 - \sin 20)}{- \sqrt{3} \sin 20 \cos 20}$
$ =\dfrac{ \sqrt{3} \cos 20- \sin 20}{\sqrt{3} \sin 20 \cos 20}$
(Multiply & Divide in Numerator & denominator by $2$ we get.  )
$=\dfrac{2 \left( \dfrac{\sqrt{3}}{2} \cos 20- \dfrac{1}{2} \sin 20  \right)}{\dfrac{\sqrt{3}}{2} (2 \sin 20 \cos 20)}$
$=\dfrac{2 (\sin 60 \cos 20- \cos 60 \sin 20)}{\dfrac{\sqrt{3}}{2} (\sin 40)}$
$=\dfrac{4}{ \sqrt{3}} \dfrac{\sin (60-20)}{\sin (40)}=\dfrac{4}{\sqrt{3}}= \dfrac{4\sqrt{3}}{3} $
So, value is $4 \sqrt{3}/3$
Multiple choice mathematics and statistics angle and its measurement degree measure of angle measure of angle radians or degrees

Convert $40^\circ \,20'$ into radian measure.

  1. $\dfrac {121}{540}\pi $ radians
  2. $\dfrac {121}{570}\pi $ radians
  3. $\dfrac {120}{513}\pi $ radians

  4. None
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given: ${40^0}$${{{20}^{'}}}$

${40^0} + \dfrac{{{{20}^0}}}{{{{60}^0}}}  $

$=40 + \dfrac{1^o}{3} = \dfrac{{{{121}^0}}}{3}$

$radian = \dfrac{\pi }{{180^o}} \times \dfrac{{121^o}}{3}$

$\boxed{ = \dfrac{{121}}{{540}}\pi \;radians}$
Multiple choice telescopic summation for infinte series binomial theorem, sequence and series maths

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$ is equal to

  1. $22$
  2. $23$
  3. $44$
  4. $45$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Now,

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\sin^2 46^o+... +\sin^2 98^o)+ \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\cos^2 44^o+... +\cos^2 2^o)+ 1$ [ Since $\sin^2 x^o=\cos^2 (90^o-x^o)$]
$=(\sin^2 2^o+\cos^2 2^o)+(\sin^2 4^o+\cos^2 4^o)+......+(\sin^2 44^o+\cos^2 44^o)+1$
$=1+1+.....+1(22\text{th})+1$
$=22+1$
$=23$.

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

The value of sin $ 2^o $ is approximately

  1. $ 2^o $
  2. $0.035$
  3. $ \frac {\pi}{180} $
  4. $0.017$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For small angles in radians, sin(x) is approximately x. 2 degrees = 2 * (pi/180) radians = pi/90 radians. pi/90 is approximately 3.14159 / 90 = 0.0349, which rounds to 0.035.

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

Value of $ \displaystyle \sin 45^{\circ} \cos 45 \left ( \tan 45^{\circ}+\cot 45^{\circ} \right )^{2}   $  is 

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

$ \displaystyle \sin 45^{\circ} \cos 45 \left ( \tan 45^{\circ}+\cot 45^{\circ} \right )^{2}   $

$=\dfrac{1}{\sqrt2} \times \dfrac{1}{\sqrt2} (1+1)^2 $


$=\dfrac42$

$=2$