Question

Trignometry Challenge , let's see who does first (20 points )
if sinQ + cosQ = 1
then prove 2sinQ - cosQ °= 2

Answer 1

sinQ+cosQ=1

we know ,
sin^2Q +cos^2Q=1
(sinQ +cosQ)^2-2sinQ.cosQ=1
1^2-2sinQ.cosQ=1
sinQ.cosQ=0
hence,
sinQ=0 or cosQ=0
again ,
sinQ-cosQ= root {(sinQ+cosQ)^2-4sinQ.cosQ}=root {(1)^2-0}= +_1

if sinQ-cosQ=1
then sinQ=1 and cosQ=0

but when sinQ-cosQ=-1
then sinQ=0 and cosQ=1
but we have given to proved
2sinQ-cosQ=2
so, here we choose only sinQ=1 and cosQ=0

now you see when sinQ=1 and cosQ=0
then 2sinQ-cosQ=2 possible.

Answer 2

${\huge {\overbrace {\underbrace {\red{answer}}}}}$

Given:-

sin^2Q +cos^2Q=1

(sinQ +cosQ)^2-2sinQ.cosQ=1

1^2-2sinQ.cosQ=1

sinQ.cosQ=0

Solution:-

♠sinQ=0 or cosQ=0

♠sinQ-cosQ= root {(sinQ+cosQ)^2-4sinQ.cosQ}=root {(1)^2-0}= +_1

sinQ-cosQ=1

therefore,sinQ=1 and cosQ=0

♠ when sinQ-cosQ=-1

♠sinQ=0 and cosQ=1

Proof:-

2sinQ-cosQ=2

2sinQ-cosQ=2sinQ=1 and cosQ=0

verified:-

$\color{red}\boxed{sinQ=1}$

$\color{red}\boxed{cosQ=0}$

$\color{red}\boxed{2sinQ-codQ=2}$

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