Given the function f(θ)= cos2θ − sinθ. (for 0°≤θ≤π) plot the graph for intervals of (π/6). hence find the value of cos2θ=sinθ.


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Question Number 47586 by Rio Michael last updated on 11/Nov/18

$G i v e n t h e f u n c t i o n f (θ) = c o s 2 θ - s i n θ . (f o r 0 ° ⩽ θ ⩽ π)$ $p l o t t h e g r a p h f o r i n t e r v a l s o f \frac{π}{6} .$ $h e n c e f i n d t h e v a l u e o f c o s 2 θ = s i n θ .$
	Answered by Joel578 last updated on 12/Nov/18

	Commented by Joel578 last updated on 12/Nov/18

	$\cos 2 θ = \sin θ$ $1 - {2 \sin}^{2} θ = \sin θ$ ${2 \sin}^{2} θ + \sin θ - 1 = 0$ $(2 \sin θ - 1) (\sin θ + 1) = 0$ $\sin θ = \frac{1}{2} \lor \sin θ = - 1$
	Commented by Rio Michael last updated on 13/Nov/18

	$t h a n k s s i r . B u t i t h o u g h t$ $c o s 2 θ = s i n θ$ $c o u l d b e l o c a t e d f r o m t h e g r a p h . T h a t i s a t p o i n t w h e r e$ $c o s 2 θ - s i n θ (f (θ)) = 0$
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