Trigonometric Identities and Equations

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.

$x+\dfrac { 1 }{ x } =2\cos { \theta  } \quad & \quad y+\dfrac { 1 }{ y } =2\cos { \phi  } \ $

$\Rightarrow x=\cos { \theta  } +i\sin { \theta  } =cis\theta \ \quad & \quad y=\cos { \phi  } +i\sin { \phi  } =cis\phi \ $

$\dfrac { x }{ y } +\dfrac { y }{ x } =\dfrac { cis\theta  }{ cis\phi  } +\dfrac { cis\phi  }{ cis\theta  } =cis\left( \theta -\phi  \right) +cis\left( -\theta +\phi  \right) $

$\therefore \quad \dfrac { x }{ y } +\dfrac { y }{ x } =2\cos { \left( \theta -\phi  \right)  } $

${ x }^{ m }{ y }^{ n }={ \left( cis\theta  \right)  }^{ m }{ \left( cis\phi  \right)  }^{ n }=\left( cism\theta  \right) \left( cisn\phi  \right) $          ...{ De Moivre's Theorem}

$\therefore \quad { x }^{ m }{ y }^{ n }=cis\left( m\theta +n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } +i\sin { \left( m\theta +n\phi  \right)  } $

${ x }^{ -m }{ y }^{ -n }={ \left( cis\theta  \right)  }^{ -m }{ \left( cis\phi  \right)  }^{ -n }=\left( cis\left( -m\theta  \right)  \right) \left( cis\left( -n\phi  \right)  \right) \ \therefore \quad { x }^{ -m }{ y }^{ -n }=cis\left( -m\theta -n\phi  \right) =\cos { \left( m\theta +n\phi  \right)  } -i\sin { \left( m\theta +n\phi  \right)  } \ $

$\therefore \quad { x }^{ m }{ y }^{ n }+{ x }^{ -m }{ y }^{ -n }=2\cos { \left( m\theta +n\phi  \right)  } $

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.

The value of the sine of 30 degrees according to the Surya Siddhanta is 0.5.

The double-angle identity for sine states that the sine of twice an angle is equal to twice the sine of the angle multiplied by the cosine of the angle.

The double-angle identity for cosine states that the cosine of twice an angle is equal to the square of the cosine of the angle minus the square of the sine of the angle.

The half-angle identity for sine states that the sine of half an angle is equal to the square root of one minus the cosine of the angle divided by two.

The half-angle identity for cosine states that the cosine of half an angle is equal to the square root of one plus the cosine of the angle divided by two.

The sum-to-product identity for sine states that the sum of two sines is equal to twice the sine of half the sum of the angles multiplied by the cosine of half the difference of the angles.

The sum-to-product identity for cosine states that the sum of two cosines is equal to twice the cosine of half the sum of the angles multiplied by the cosine of half the difference of the angles.

The product-to-sum identity for sine states that the product of two sines is equal to half the difference of the cosines of the sum and difference of the angles.

The product-to-sum identity for cosine states that the product of two cosines is equal to half the sum of the cosines of the sum and difference of the angles.