Question

Prove that sin3a-cos3a/sina+cosa=2sin2a-1

Answer 1

`(sin(3a)-cos(3a))/(sin(a)+cos(a)) =2sin(2a)-1`

Explanation:

we are given

`(sin(3a)-cos(3a))/(sin(a)+cos(a)) =2sin(2a)-1`

we can simplify left side and make it equal to right side

we can use trig identity

`sin(3a)=3sin(a)-4sin^3(a)`

`cos(3a)=4cos^3(a)-3cos(a)`

now, we can plug values

`((3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a)))/(sin(a)+cos(a))`

now, we can simplify

`(3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a))/(sin(a)+cos(a))`

`(3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a))/(sin(a)+cos(a))`

`(3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a)))/(sin(a)+cos(a))`

now, we can factor it

`(3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a))/(sin(a)+cos(a))`

`((sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)])/(sin(a)+cos(a))`

we can use trig identity

`sin^2(a)+cos^2(a)=1`

`((sin(a)+cos(a))[3-4(1-sin(a)cos(a)])/(sin(a)+cos(a))`

we can cancel terms

`=3-4(1-sin(a)cos(a))`

now, we can simplify it further

`=3-4+4sin(a)cos(a))`

`=-1+4sin(a)cos(a))`

`=4sin(a)cos(a)-1`

`=2* 2sin(a)cos(a)-1`

now, we can use trig identity

`2sin(a)cos(a)=sin(2a)`

we can replace it

`=2sin(2a)-1`

so,

`(sin(3a)-cos(3a))/(sin(a)+cos(a)) =2sin(2a)-1`

Answer 2

To prove : sin(3a)-cos(3a)/sin(a)+cos(a) = 2sin(2a)-1

Explanation:

\frac{\sin3a-\cos3a}{\sin a+\cos a}=2\sin 2a-1\\\cos3a=4\cos^{3}a-3\cos a\\\sin 3a=3\sin a-4\sin ^{3}a\\substituting the value, we get\\L.H.S= \frac{3\sin a-4\sin ^{3}a-[4\cos ^{3}a-3\cos a]}{\sin a+\cos a}\\\frac{3(\sin a+\cos a)-4(\sin ^{3}a+4\cos ^{3}a)}{\sin a+\cos a}\\\frac{3(\sin a+\cos a)-4[(\cos a+\sin a)(\cos ^{2}a+\sin ^{2}a-\sin a\cos a)]}{\sin a+\cos a}\\\frac{3(\sin a+\cos a)-4[(\cos a+\sin a)(1-\sin a\cos a)]}{\sin a+\cos a}\\\frac{(\sin a+\cos a)[3-4(1-\sin a\cos a)]}{\sin a+\cos a}\\\frac{(\sin a+\cos a)[3-4+4\sin a\cos a]}{\sin a+\cos a}\\\frac{(\sin a+\cos a)(-1+4\sin a\cos a)}{\sin a+\cos a}\\-1+4\sin a\cos a\\-1+2\sin 2a\\2\sin 2a-1=R.H.S

Since, sin(2a)= 2sin(a)cos(a)

2sin(2a)= 4sin(a)cos(a)