If cos 40^circ = x and cos theta = 1 - 2x^2,t | vedclass

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Question

(A) Given $\cos 	heta = 1 - 2x^2$ and $x = \cos 40^\circ$. Substituting $x$,we get $\cos 	heta = 1 - 2\cos^2 40^\circ$. Using the identity $\cos 2A

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$100^\circ$ and $260^\circ$

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(A) Given $\cos 	heta = 1 - 2x^2$ and $x = \cos 40^\circ$. Substituting $x$,we get $\cos 	heta = 1 - 2\cos^2 40^\circ$. Using the identity $\cos 2A = 2\cos^2 A - 1$,we have $\cos 	heta = -(2\cos^2 40^\circ - 1) = -\cos(2 	imes 40^\circ) = -\cos 80^\circ$. To find $	heta$ in the range $[0^\circ, 360^\circ]$,we use the property $\cos(180^\circ - A) = -\cos A$ and $\cos(180^\circ + A) = -\cos A$. Thus,$\cos 	heta = \cos(180^\circ - 80^\circ) = \cos 100^\circ$ and $\cos 	heta = \cos(180^\circ + 80^\circ) = \cos 260^\circ$. Therefore,the possible values of $	heta$ are $100^\circ$ and $260^\circ$.

If $\cos 40^\circ = x$ and $\cos \theta = 1 - 2x^2$,then the possible values of $\theta$ lying between $0^\circ$ and $360^\circ$ are:

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