We have, #10; #10; sin( θ +π6) cos( #10;θ +π3) #10; #10; =sinθcosπ6+cosθsinπ #10;6 cosθcosπ3+sinθsinπ3 #10; #10; =√(3)2sinθ +12cos ...

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Sign of Trigonometric Ratios in Different Quadrants

sin θ +π 6 -c...

Question

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

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Solution

We have,

$sin (θ + \frac{π}{6}) - cos (θ + \frac{π}{3})$

$= sin θ cos \frac{π}{6} + cos θ sin \frac{π}{6} - cos θ cos \frac{π}{3} + sin θ sin \frac{π}{3}$

$= \frac{\sqrt{3}}{2} sin θ + \frac{1}{2} cos θ - \frac{1}{2} cos θ + \frac{\sqrt{3}}{2} sin θ$

$= 2 \frac{\sqrt{3}}{2} sin θ$

$= \sqrt{3} sin θ$
Hence, this is the answer.

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Similar questions

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

I f 2 sin (θ + \frac{π}{3}) = cos (θ - \frac{π}{6}), t h e n tan θ + \sqrt{3} = 0

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}

\frac{d}{d θ} ({tan}^{- 1} (\frac{1 - cos θ}{sin θ}))

- π < θ < π

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