Prove that sin - - 6 -cos - - 3 - 3 -sin-

sin ( θ π6 )+cos ( θ π3 ) = ( sinθcosπ6 sinπ6cosθ #160; #160; )+ ( cosθcosπ3+sinθsinπ3 ) =sinθ√(3)2 12cosθ +12cosθ +sinθ√(3)2 =2×√(3) ...

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Sin(A+B)Sin(A-B)

Prove that s...

Question

Prove that
$sin (θ - \frac{π}{6}) + cos (θ - \frac{π}{3}) = \sqrt{3} . sin θ$

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s i n (θ + \frac{π}{6}) - c o s (θ + \frac{π}{3})

\sqrt{\frac{1 - sin θ}{1 + sin θ}} + \sqrt{\frac{1 + sin θ}{1 - sin θ}}

= \frac{- 2}{cos θ}, \frac{π}{2} < θ < π

If $m = (c o s θ - s i n θ)$ and $n = (c o s θ + s i n θ)$ then show that

$\sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{2}{\sqrt{1 - t a n^{2} θ}}$

\begin{matrix} X = sin (θ + \frac{7 π}{12}) + sin (θ - \frac{π}{12}) + sin (θ + \frac{3 π}{12}), Y = cos (θ - \frac{7 π}{12}) + cos (θ - \frac{π}{12}) + cos (θ + \frac{3 π}{12}) p r o v e t h a t \frac{x}{y} - \frac{y}{x} = 2 tan 2 θ . \end{matrix}

t a n (\begin{matrix} \frac{π}{2} s i n θ \end{matrix}) = c o t (\begin{matrix} \frac{π}{2} c o s θ \end{matrix})

s i n θ + c o s θ = \pm \sqrt{2}

- \sqrt{2} \leq s i n θ + c o s θ \leq \sqrt{2}

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