Question

if sinx + cosx =17/13 
then sinx-cosx=

Answer 1

squaring on both sides
sin²2A+cos²2A+2sinAcosA=289/169
1+2sinAcosA=289/169
2sinAcosA=289/169-1
2sinAcosA=120/169
sinAcosA=60/169
now
sinA-cosA=root(sinA-cosA)²
sinA-cosA=root(sin²2A+cos²2A-2sinAcosA)
sinA-cosA=root(1-2SinacosA)
sinA-cosA=root(1-2*60/169)
sinA-cosA=root(49/169)
sinA-cosA=7/13
hence the answer of sinA-cosA=7/13

Answer 2

Answer:

$sinx-cosx=\frac{7}{13}$

Step-by-step explanation:

Given : $sinx + cosx =\frac{17}{13}$

To Find : sinx - cos x

Solution :

$sinx + cosx =\frac{17}{13}$

Squaring on both sides

$sin^2x+cos^2x+2sinxcosx=\frac{289}{169}$

Identity : $sin^2x +Cos^2x = 1$

$1+2sinxcosx=\frac{289}{169}$

$2sinxcosx=\frac{289}{169}-1$

$2sinxcosx=\frac{120}{169}$

$sinxcosx=\frac{60}{169}$

we can rewrite

$sinx-cosx=\sqrt{(sinx-cosx)^2}$

$sinx-cosx=\sqrt{sin^2x+cos^2x-2sinxcosx}$

$sinx-cosx=\sqrt{1-2sinxcosx}$

$sinx-cosx=\sqrt{1-\frac{120}{169}}$

$sinx-cosx=\sqrt{\frac{49}{169}}$

$sinx-cosx=\frac{7}{13}$

Hence $sinx-cosx=\frac{7}{13}$