Question

if cos θ + sin θ = √2 cos θ, then the value of Cos θ - Sin θ=

Answer 1

Answer:

Let θ = x

cos x + sin x = √2 cos x

squaring on both side, we get......

cos2x + sin2x + 2cosxsinx = 2cos2x

2sinxcosx = 2cos2x - cos2x - sin2x

2sinxcosx = cos2x - sin2x

2sinxcosx = (cosx+sinx) (cosx - sinx)

2sinxcosx = (root2 cosx) (cosx - sinx)

2sinxcosx/root2 cosx = cosx - sinx

√2 sinx = cosx - sinx

Answer 2

Given:-

• $\sf \: cos( \theta) + sin( \theta) = \: \sqrt{2} cos( \theta)$

To find:-

• $\sf \: cos( \theta) - sin( \theta)$

Answer:-

Given that,

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

Squaring both sides, we get,

$\longrightarrow \: \sf \: \Big \{cos( \theta) + sin( \theta) \Big \}^{2} = \Big \{ \sqrt{2} cos( \theta) \Big \} ^{2}$

• Using (a + b)² = a² + b² + 2ab in L.H.S.,

$\longrightarrow \sf \: {cos}^{2} ( \theta) + {sin}^{2} ( \theta) + 2sin( \theta)cos( \theta) = 2 {cos}^{2} ( \theta)$

$\longrightarrow \sf 2sin( \theta)cos( \theta) = 2 {cos}^{2} ( \theta) - {cos}^{2} ( \theta) - {sin}^{2} ( \theta)$

$\longrightarrow \sf 2sin( \theta)cos( \theta) = {cos}^{2} ( \theta) - {sin}^{2} ( \theta)$

• Using a² - b² = (a + b)(a - b) in R.H.S.,

$\longrightarrow \sf 2sin( \theta)cos( \theta) = \Big \{ cos( \theta) - sin( \theta)\Big\} \Big\{cos( \theta) + sin( \theta)\Big\}$

• As it is given that cos(θ) + sin(θ) = √2 sin(θ),

$\longrightarrow \sf 2sin( \theta)cos( \theta) = \Big\{cos( \theta) - sin( \theta)\Big\} \Big\{ ( \sqrt{2} cos( \theta)\Big\}$

$\longrightarrow \sf cos( \theta) - sin( \theta) = \dfrac{2sin( \theta)cos( \theta)}{ \sqrt{2}cos( \theta) }$

$\longrightarrow \sf cos( \theta) - sin( \theta) = \dfrac{ { \cancel{2}}^ {{ }^{ \sqrt{2} }} sin( \theta) \cancel{cos( \theta)}}{ \cancel{ \sqrt{2}} \cancel{cos( \theta)} }$

$\longrightarrow \sf cos( \theta) \ - sin( \theta) = \sqrt{2} sin( \theta)$

Hence the answer is,

$\longrightarrow \underline{\underline{\sf{\green{ cos(\theta) - sin(\theta) = \sqrt{2} sin(\theta) }}}}$