Question

If cos thetha + sin thetha = √2 cos thetha, prove that cos thetha -sin thetha = +_ √2 sin thetha​

Answer 1

Answer:

tan(theta)=coefficient of friction (u) tan(theta) = √3 ... Find theta for which sintheta = costheta, if 180° < 0 < 360 °(1) ...

Answer 2

$\rm{\cos \theta+\sin \theta=\sqrt{2} \cos \theta}$

• Squaring On Both Sides of The Equation

$\to\rm{(\cos \theta+\sin \theta)^2=(\sqrt{2} \cos \theta)^2}$

• Apply Algebraic Identity : $\rm{(a+b)^2=a^2+b^2+2ab}$

$\to\rm{\cos^2 \theta+\sin^2 \theta+2\times \cos\theta\times \sin\theta=(\sqrt{2} \cos \theta)^2}$

$\to\rm{\cos^2 \theta+\sin^2 \theta+2\sin\theta\cos\theta=(\sqrt{2} \cos \theta)^2}$

$\to\rm{\cos^2 \theta+\sin^2 \theta+2\sin\theta\cos\theta=2 \cos^2 \theta}$

• Bring cos²θ and sin²θ to Left Sides of The Equation

$\to\rm{2\sin\theta\cos\theta=2 \cos^2 \theta - \cos^2 \theta-\sin^2 \theta}$

$\to\rm{2\sin\theta\cos\theta=\cos^2 \theta-\sin^2 \theta}$

• Apply Difference of Two Products Rule : $\rm{a^2-b^2=(a+b)(a-b)}$

$\to\rm{2\sin\theta\cos\theta=(\cos \theta+\sin \theta)(\cos \theta-\sin \theta)}$

• We are already given With : $\rm{\cos \theta+\sin \theta=\sqrt{2} \cos \theta}$

$\to\rm{2\sin\theta\cos\theta=\sqrt{2}\cos\theta(\cos \theta-\sin \theta)}$

• Dividing Both Side by $\rm{\sqrt{2}\cos\theta}$

$\to\rm{\dfrac{2\sin\theta\cos\theta}{\sqrt{2}\cos\theta}=\dfrac{\sqrt{2}\cos\theta(\cos \theta-\sin \theta)}{\sqrt{2}\cos\theta}}$

$\to\rm{\dfrac{2\sin\theta}{\sqrt{2}}=\dfrac{\cos \theta-\sin \theta}{1}}$

$\to\rm{\dfrac{2\sin\theta}{\sqrt{2}}=\cos \theta-\sin \theta}$

• Switch Sides

$\to\rm{\cos \theta-\sin \theta=\dfrac{2\sin\theta}{\sqrt{2}}}$

• We know a = √a × √a , So : $\rm{2=\sqrt{2}\times\sqrt{2}}$

$\to\rm{\cos \theta-\sin \theta=\dfrac{\sqrt{2}\times\sqrt{2} \sin\theta}{\sqrt{2}}}$

• Cancelling √2 in Numerator and Denominator

$\to\rm{\cos \theta-\sin \theta=\sqrt{2} \sin\theta}$

Hence Proved !!!!!