Respuesta :
Answer:
[tex]\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1[/tex]
Step-by-step explanation:
we are given
[tex]\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1[/tex]
we can simplify left side and make it equal to right side
we can use trig identity
[tex]sin(3a)=3sin(a)-4sin^3(a)[/tex]
[tex]cos(3a)=4cos^3(a)-3cos(a)[/tex]
now, we can plug values
[tex]\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)} [/tex]
now, we can simplify
[tex]\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)} [/tex]
[tex]\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)} [/tex]
[tex]\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)} [/tex]
now, we can factor it
[tex]\frac{3(sin(a)+cos(a))-4(sin(a)+cos(a))(sin^2(a)+cos^2(a)-sin(a)cos(a)}{sin(a)+cos(a)} [/tex]
[tex]\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)} [/tex]
we can use trig identity
[tex]sin^2(a)+cos^2(a)=1[/tex]
[tex]\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)} [/tex]
we can cancel terms
[tex]=3-4(1-sin(a)cos(a))[/tex]
now, we can simplify it further
[tex]=3-4+4sin(a)cos(a))[/tex]
[tex]=-1+4sin(a)cos(a))[/tex]
[tex]=4sin(a)cos(a)-1[/tex]
[tex]=2\times 2sin(a)cos(a)-1[/tex]
now, we can use trig identity
[tex]2sin(a)cos(a)=sin(2a)[/tex]
we can replace it
[tex]=2sin(2a)-1[/tex]
so,
[tex]\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1[/tex]
Answer:
To prove : sin(3a)-cos(3a)/sin(a)+cos(a) = 2sin(2a)-1
Step-by-step 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)
