Tag: trigonometric ratios of some specific angles

Questions Related to trigonometric ratios of some specific angles

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

Values of : $sin{ 10 }^{ 0 }sin{ 50 }^{ 0 }sin{ 60 }^{ 0 }sin{ 70 }^{ 0 }$ is

  1. $\cfrac { 3 }{ 16 } $
  2. $\cfrac { 5 }{ 16 } $
  3. $\cfrac { \sqrt { 3 } }{ 16 } $
  4. $\cfrac { \sqrt { 5 } }{ 16 } $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

As we know that

$\sin A\sin(60^{\circ}-A)\sin (60^{\circ}+A)=\dfrac{1}{4}\sin 3 A$
Put $A=10^{\circ}$
So $\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\dfrac{1}{4}\sin 30^{\circ}=\dfrac{1}{8}$
So $\sin 10^{\circ}\sin 60^{\circ}\sin 50^{\circ}\sin 70^{\circ}=\sin 60^{\circ}(\sin 10^{\circ}\sin 50^{\circ}\sin 70^{\circ})=\dfrac{\sqrt{3}}{2}\times \dfrac{1}{8}=\dfrac{\sqrt{3}}{16}$

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

The value of $(4 \, cos^2 9^o - 1) (4 cos^2 27^o - 1) (4 \, cos^2 81^o - 1) (4 \, cos^2 243^o - 1) $ is 

  1. 1

  2. -1

  3. 2

  4. none of these

Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a product of the form (4cos^2(x)-1). Using the identity 4cos^2(x)-1 = sin(3x)/sin(x), the product telescopes to sin(3^n * x) / (sin(x) * 2^n). For the given angles, the result is 1.

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

Find the value of, $\dfrac {4}{3}\cot^{2}30^{o}+\cot^{2}60^{o}-2\csc ^{2}60^{o}-\dfrac {3}{4}\tan^{2}30^{o}$

  1. $10/3$
  2. $11/3$
  3. $4$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\dfrac { 4 }{ 3 } { \cot }^{ 2 }30+{ \cot }^{ 2 }60-2{ csc }^{ 2 }60-\dfrac { 3 }{ 4 } { \tan }^{ 2 }30$

$\Rightarrow \dfrac { 4 }{ 3 } { \left( \sqrt { 3 }  \right)  }^{ 2 }+{ \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  }^{ 2 }-2\times { \left( \dfrac { 2 }{ \sqrt { 3 }  }  \right)  }^{ 2 }-\dfrac { 3 }{ 4 } \times { \left( \dfrac { 1 }{ \sqrt { 3 }  }  \right)  }^{ 2 }$
$\Rightarrow 4+\dfrac { 1 }{ 3 } -\dfrac { 8 }{ 3 } -\dfrac { 1 }{ 4 } $
$\Rightarrow \dfrac { 15 }{ 4 } -\dfrac { 7 }{ 3 } $
$=\dfrac { 17 }{ 12 } $
None of these.