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Question: If \(\cos\theta = 0\) and \(\sin^{2}\theta = \frac{1}{2} = \sin^{2}\left( \frac{\pi}{4} \right)\), t...

If cos⁡θ=0\cos\theta = 0 and sin⁡2θ=12=sin⁡2(π4)\sin^{2}\theta = \frac{1}{2} = \sin^{2}\left( \frac{\pi}{4} \right), then the possible values of ⇒\Rightarrowlying between θ=(2n+1)π2\theta = (2n + 1)\frac{\pi}{2}and θ=nπ±π4\theta = n\pi \pm \frac{\pi}{4}is .

A

⇒\Rightarrowand θ=90o\theta = 90^{o}

B

45o45^{o}andsin⁡(θ+π6)=1=sin⁡(π2)⇒θ=π2−π6=π3\sin\left( \theta + \frac{\pi}{6} \right) = 1 = \sin\left( \frac{\pi}{2} \right) \Rightarrow \theta = \frac{\pi}{2} - \frac{\pi}{6} = \frac{\pi}{3}

C

cos⁡Asin⁡(A−π6)=12[sin⁡(2A−π6)−sin⁡π6]\cos A\sin\left( A - \frac{\pi}{6} \right) = \frac{1}{2}\left\lbrack \sin\left( 2A - \frac{\pi}{6} \right) - \sin\frac{\pi}{6} \right\rbrackand sin⁡(2A−π6)−12\sin\left( 2A - \frac{\pi}{6} \right) - \frac{1}{2}

D

2A−π6=π22A - \frac{\pi}{6} = \frac{\pi}{2}and ⇒A=π3\Rightarrow A = \frac{\pi}{3}

Answer

⇒\Rightarrowand θ=90o\theta = 90^{o}

Explanation

Solution

Here a=bcos⁡C+ccos⁡Ba = b\cos C + c\cos B = sin⁡Bsin⁡(A+B)=sin⁡Bsin⁡C=bc\frac{\sin B}{\sin(A + B)} = \frac{\sin B}{\sin C} = \frac{b}{c}

sin⁡(A−B)sin⁡(A+B)=sin⁡Acos⁡B−sin⁡Bcos⁡Asin⁡C\frac{\sin(A - B)}{\sin(A + B)} = \frac{\sin A\cos B - \sin B\cos A}{\sin C}= =accos⁡B−bccos⁡A= \frac{a}{c}\cos B - \frac{b}{c}\cos A

= cos⁡B=a2+c2−b22ac,cos⁡A=b2+c2−a22bc\cos B = \frac{a^{2} + c^{2} - b^{2}}{2ac},\cos A = \frac{b^{2} + c^{2} - a^{2}}{2bc}

Hence, ⇒accos⁡B−bccos⁡A=12c2\Rightarrow \frac{a}{c}\cos B - \frac{b}{c}\cos A = \frac{1}{2c^{2}} i.e., (a2+c2−b2−b2−c2+a2)(a^{2} + c^{2} - b^{2} - b^{2} - c^{2} + a^{2}) and 260°.