Question

13. If cose theta + sin theta root2 cose theta then prove that cos theta - sin thetq = root 2 sin​ theta

Answer 1

Question:

If $\displaystyle{\sf\:\cos\:\theta\:+\:\sin\:\theta\:=\:\sqrt{2}\:\cos\:\theta}$, then prove that, $\displaystyle{\sf\:\cos\:\theta\:-\:\sin\:\theta\:=\:\sqrt{2}\:\sin\:\theta}$.

Answer:

$\displaystyle{\boxed{\red{\sf\:\cos\:\theta\:-\:\sin\:\theta\:=\:\sqrt{2}\:\sin\:\theta}}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\cos\:\theta\:+\:\sin\:\theta\:=\:\sqrt{2}\:\cos\:\theta}$

We have to prove that,

$\displaystyle{\sf\:\cos\:\theta\:-\:\sin\:\theta\:=\:\sqrt{2}\:\sin\:\theta}$

Now,

$\displaystyle{\sf\:\cos\:\theta\:+\:\sin\:\theta\:=\:\sqrt{2}\:\cos\:\theta}$

$\displaystyle{\implies\sf\:(\:\cos\:\theta\:+\:\sin\:\theta\:)^2\:=\:(\:\sqrt{2}\:\cos\:\theta\:)^2\:\:\:-\:-\:-\:[\:Taking\:square\:on\:both\:sides\:]}$

$\displaystyle{\implies\sf\:\cos^2\:\theta\:+\:2\:\times\:\cos\:\theta\:\times\:\sin\:\theta\:+\:\sin^2\:\theta\:=\:\sqrt{2}\:\times\:\sqrt{2}\:\times\:\cos^2\:\theta}$

$\displaystyle{\implies\sf\:\cos^2\:\theta\:+\:2\:\cos\:\theta\:\sin\:\theta\:+\:\sin^2\:\theta\:=\:2\:\cos^2\:\theta}$

$\displaystyle{\implies\sf\:2\:\cos\:\theta\:\sin\:\theta\:=\:2\:\cos^2\:\theta\:-\:\cos^2\:\theta\:-\:\sin^2\:\theta}$

$\displaystyle{\implies\sf\:2\:\cos\:\theta\:\sin\:\theta\:=\:\cos^2\:\theta\:-\:\sin^2\:\theta}$

$\displaystyle{\implies\sf\:\cos^2\:\theta\:-\:\sin^2\:\theta\:=\:2\:\cos\:\theta\:\sin\:\theta}$

$\displaystyle{\implies\sf\:(\:\cos\:\theta\:+\:\sin\:\theta\:)\:(\:\cos\:\theta\:-\:\sin\:\theta\:)\:=\:2\:\cos\:\theta\:\sin\:\theta\:\:\:-\:-\:\left[\:a^2\:-\:b^2\:=\:(\:a\:+\:b\:)\:(\:a\:-\:b\:)\:\right]}$

$\displaystyle{\implies\sf\:\sqrt{2}\:\cos\:\theta\:\times\:\left(\:\cos\:\theta\:-\:\sin\:\theta\:\right)\:=\:2\:\cos\:\theta\:\sin\:\theta\:\:\:-\:-\:[\:Given\:]}$

$\displaystyle{\implies\sf\:\cos\:\theta\:-\:\sin\:\theta\:=\:\dfrac{\cancel{2}\:\cancel{\cos\:\theta}\:\sin\:\theta}{\cancel{\sqrt{2}}\:\cancel{\cos\:\theta}}}$

$\displaystyle{\implies\boxed{\red{\sf\:\cos\:\theta\:-\:\sin\:\theta\:=\:\sqrt{2}\:\sin\:\theta}}}$

Hence proved!

Answer 2

Question :

$if \:( cosθ - sinθ )= \sqrt{2} cosθ \: \: then \: \:$

$then \: prove \: that \: (cosθ - sinθ) = \sqrt{2}sinθ$

To Prove

$prove \: \: that \: \: (cosθ - sinθ) = \sqrt{2}sinθ$

Answer :

Now,

$cosθ + sinθ = \sqrt{2} cosθ$

Squaring on both sides,

${(cos θ + sinθ )^{2} } = 2 {cos}^{2} θ$

${cos}^{2} θ + {sin}^{2} θ + 2cosθsinθ = 2 {cos}^{2} θ$

$2cosθsinθ = 2 {cos}^{2} θ - {cos}^{2} θ - {sin}^{2} θ$

$2cosθsinθ = {cos}^{2} θ - {sin}^{2} θ$

Since,

$( {a}^{2} - {b}^{2} ) = (a + b)(a - b)$

$2cosθsinθ = (cosθ + sinθ)(cosθ - sinθ)$

We know that,

$(cosθ + sinθ) = \sqrt{2} cosθ$

hence,

$2cosθsinθ = ( \sqrt{2} sinθ)(cosθ - sinθ)$

$(cos θ -sinθ) = \frac{ \sqrt{2} \times \sqrt{2} \times cos θ \times sinθ}{ \sqrt{2} \times cosθ}$

$(cosθ \: - \: sinθ) \: = \: \sqrt{2} sinθ$

Hence proved.