Question

If
Sin∅ + cos ∅ = √2 cos∅
Then prove
Cos∅ - sin∅ = √2 sin ∅

Answer 1

HEY MATE YOUR ANSWER IS HERE

★ GIVEN ★

SIN∅ + COS∅ = √2 COS∅

★ TO PROVE ★

COS∅ - SIN∅ = √2 SIN∅

★ SOLUTION ★

→ SIN∅ + COS∅ = √2 COS∅

→ SIN∅ = √2 COS∅ - COS∅

→ SIN∅ = COS∅ ( √2 - 1 )

→ (SIN∅)/(√2 - 1 ) = COS∅

NOW BY RATIONALISING WE GET

$\frac{sin \: </strong><strong>}{ \sqrt{2} - 1 \: } \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1} = cos </strong><strong>\\ \\ \frac{sin </strong><strong>\: ( \sqrt{2} + 1 )}{2 - 1} = cos </strong><strong>\\ \\ \frac{</strong><strong>sin</strong><strong> \: ( \sqrt{2} + 1 )}{1} = </strong><strong>cos</strong><strong> \\ \\ \sqrt{2} </strong><strong>sin</strong><strong> \: + \: </strong><strong>sin</strong><strong> \: = \: cos </strong><strong>\\ \\ \sqrt{2 \:} </strong><strong>sin</strong><strong> \: = </strong><strong>cos</strong><strong>\: - \: </strong><strong>sin</strong><strong> \: \\ \\ hence \: proved$

THANKS FOR UR QUESTION HOPE IT HELPS

Answer 2

Answer:

Given sin theta + cos theta = sqrt(2)cos theta prove cos theta - sin theta = sqrt(2)sin theta

sin theta + cos theta = sqrt(2)cos theta square both sides

sin^2theta + cos^2 theta + 2sintheta costheta=2cos^2theta

sin^2theta - cos^2theta + 2sinthetacostheta=0

-sin^2theta + cos^2theta -2sinthetacostheta=0 Add 2sin^2theta to both sides

sin^2theta+cos^2theta-2sinthetacostheta=2sin^2theta

(costheta-sintheta)^2=2sin^2theta

costheta-sintheta=sqrt(2)sintheta as required.