Trigonometric Identities and Equations

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.

Brewster's angle i_p satisfies tan(i_p) = mu, and critical angle theta_c satisfies sin(theta_c) = 1/mu. Thus, tan(i_p) = 1/sin(theta_c) = cosec(theta_c), so i_p = tan^-1(cosec(theta_c)).

De Moivre's theorem states (cos theta + i sin theta)^n = cos(n theta) + i sin(n theta) for integer n, and the values for fractional n are given by the corollary provided.

Let a = exp(i * 8pi/11). The sum is a + a^2 + a^3 + a^4 + a^5. This is a geometric series: a(1-a^5)/(1-a). The real part of this sum is -1/2.

By Binomial Theorem
${ \left( 1+z \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }z+{ C } _{ 2 }{ z }^{ 2 }+{ C } _{ 3 }{ z }^{ \ 3 }+....+{ C } _{ n }{ z }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }{ z }^{ r } } $      ...(1)

Substituting $z=\cos { 2\theta  } +i\sin { 2\theta  }  $ in eq. (1), we get

${ \left( 1+\cos { 2\theta  } +i\sin { 2\theta  }  \right)  }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }\left( \cos { 2\theta  } +i\sin { 2\theta  }  \right) ^{ r } } $

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\left( \cos { 2r\theta  } +i\sin { 2r\theta  }  \right)  }$      ...{De Moivre's Theorem}

$\Rightarrow { \left[ 2\cos { \theta  } \left( \cos { \theta  } +i\sin { \theta  }  \right)  \right]  }^{ n }=\sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } $

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } { \left( \cos { n\theta  } +i\sin { n\theta  }  \right)  }$         ...{De Moivre's Theorem}

$\Rightarrow \sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } +i\left( \sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  }  \right) ={ 2 }^{ n }\cos ^{ n }{ \theta  } \cos { n\theta  } +i\left( { 2 }^{ n }\cos ^{ n }{ \theta  } \sin { n\theta  }  \right) $

On comparing real and Imaginary parts, we get
$\sum _{ r=0 }^{ n }{ { C } _{ r }\cos { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } \cos { n\theta  } \quad & \quad \sum _{ r=0 }^{ n }{ { C } _{ r }\sin { 2r\theta  }  } ={ 2 }^{ n }\cos ^{ n }{ \theta  } \sin { n\theta  } $
Hence, option 'A' and 'C' are correct.

Using De-Moivre's Theorem, the given expression
$\displaystyle = \frac{\left ( \cos \theta -i\sin \theta  \right )^{14}\left ( \cos \theta +i\sin \theta  \right )^{-15}}{\left ( \cos \theta +i\sin \theta  \right )^{48}\left ( \cos \theta +i\sin \theta  \right )^{-30}}$
$\displaystyle=\frac{\left ( e^{i\theta } \right )^{-29}}{\left ( e^{i\theta } \right )^{18}}=\left ( e^{i\theta } \right )^{-47}$
$\displaystyle=\left ( \cos \theta +i\sin \theta  \right )-^{47}=\cos 47\theta -\sin47\theta.$ 

Ans: B

Changing the above expression to Eular's form, we get
$e^{i\theta}e^{2i\theta}e^{3i\theta}...e^{in\theta})=1$
$e^{i(\theta+2\theta+3\theta+...n\theta}=1$
$e^{i\cfrac{n(n+1)}{2}\theta}=e^{2m\pi}$
Therefore, simplifying we get
$\dfrac{n(n+1)}{2}\theta=2m\pi$
$\theta=\dfrac{4m\pi}{n(n+1)}$

Let
$z=(cos\theta+isin\theta)^{3}$
Taking the fifth root, we get
$z^{\dfrac{1}{5}}=(cos\theta+isin\theta)^{\dfrac{3}{5}}=x$
Now there will be $5$, corresponding values of $x$.
Product if all the $5$ values will be
$x^{5}$
$=(cos\theta+isin\theta)^{3}$
$=cos3\theta+isin3\theta$ ... using De-Moivre's rule.
Now consider $x^{n}=1$
Hence if $n$ is odd, the nth roots of unity will be
$1,a _{1},\overline{a _{1}},a _{2},\overline{a _{2}}....$
Now $a _{1}.\overline{a _{1}}=|a _{1}|^{2}=1$
$a _{2}.\overline{a _{2}}=|a _{2}|^{2}=1$
:
:
Hence one root will be one, and the rest $n-1$ roots will occur in pair with its conjugate.
Hence product will be $1$.
Substituting, $n=5$, we get the roots as
$1,a _{1},\overline{a _{1}},a _{2},\overline{a _{2}}$
Hence product if all the roots will be $1$. And sum of all the roots will be $0$.
Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

The equation cos(px) + cos(qx) = 0 implies 2cos((p+q)x/2)cos((p-q)x/2) = 0. Solving for x gives the roots in AP with the specified common differences.

Using the identity sin(A)/cos(B) related to telescoping sums, the expression simplifies. For f_2, the calculation results in -1.

We have,

$\sin \alpha ,\,{{\sin }^{2}}\alpha ,\,1,\,{{\sin }^{4}}\alpha \,\,and\,\,{{\sin }^{5}}\alpha $ in A.P.

Then,

$ \text{First}\,\text{term}\,\,a=\sin \alpha  $

$ \text{Common}\,\text{difference}\,\,\text{=}\,\text{Second}\,\text{term}\,\text{-}\,\text{first}\,\,\text{term} $

Where

$ {{T} _{1}}=\,First\,term $

$ {{T} _{2}}=Second\,term $

$ {{T} _{3}}=\,Third\,term $

$ ....... $

Then, we know that,

If the series in an A.P.

$ {{T} _{2}}-{{T} _{1}}={{T} _{3}}-{{T} _{2}}={{T} _{4}}-{{T} _{3}}={{T} _{5}}-{{T} _{4}} $

$ {{\sin }^{2}}\alpha -\sin \alpha =1-{{\sin }^{2}}\alpha ={{\sin }^{4}}\alpha -1={{\sin }^{5}}\alpha -{{\sin }^{4}}\alpha  $

$ \Rightarrow {{\sin }^{2}}\alpha -\sin \alpha =1-{{\sin }^{2}}\alpha  $

$ \Rightarrow {{\sin }^{2}}\alpha +{{\sin }^{2}}\alpha -\sin \alpha =1 $

$ \Rightarrow 2{{\sin }^{2}}\alpha -\sin \alpha -1=0 $

$ \Rightarrow 2{{\sin }^{2}}\alpha -\left( 2-1 \right)\sin \alpha -1=0 $

$ \Rightarrow 2{{\sin }^{2}}\alpha -2\sin \alpha +\sin \alpha -1=0 $

$ \Rightarrow 2\sin \alpha \left( \sin \alpha -1 \right)+1\left( \sin \alpha -1 \right)=0 $

$ \Rightarrow \left( \sin \alpha -1 \right)\left( 2\sin \alpha +1 \right)=0 $

$ \Rightarrow \sin \alpha -1=0,\,\,\,2\sin \alpha +1=0 $

$ \Rightarrow \sin \alpha =1,\,\,\,\sin \alpha =\dfrac{-1}{2} $

$ \Rightarrow \sin \alpha =\sin \dfrac{\pi }{2},\,\,\,\sin \alpha =-\sin \dfrac{\pi }{6} $

$ \Rightarrow \alpha =\dfrac{\pi }{2},\,\,\,\,\sin \alpha =\sin \left( \pi +\dfrac{\pi }{6} \right)\,\,\,\,\,\,\,\,\,\,\because \sin \left( \pi +\theta  \right)=-\sin \theta  $

$ \Rightarrow \alpha =\dfrac{\pi }{2},\,\,\,\,\,\alpha =\dfrac{7\pi }{6} $

Similarly we can show that,

$\alpha =-\dfrac{\pi }{2}$

Hence, $\alpha =\left( -\dfrac{\pi }{2},\,\dfrac{\pi }{2} \right)$

Hence, this is the required answer.

Aryabhata defined (\sin 90^\circ) to be equal to (1). This is because (\sin \theta) is the ratio of the opposite side to the hypotenuse, and in a right triangle with an angle of (90^\circ), the opposite side is equal to the hypotenuse.