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luda_lava [24]
3 years ago
11

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

Mathematics
2 answers:
makvit [3.9K]3 years ago
8 0

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)



Sloan [31]3 years ago
7 0

Answer:

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1

Step-by-step explanation:

we are given

\frac{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

\frac{(3sin(a)-4sin^3(a))-(4cos^3(a)-3cos(a))}{sin(a)+cos(a)}

now, we can simplify

\frac{3sin(a)-4sin^3(a)-4cos^3(a)+3cos(a)}{sin(a)+cos(a)}

\frac{3sin(a)+3cos(a)-4sin^3(a)-4cos^3(a)}{sin(a)+cos(a)}

\frac{3(sin(a)+cos(a))-4(sin^3(a)+cos^3(a))}{sin(a)+cos(a)}

now, we can factor it

\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)}

\frac{(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

\frac{(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\times 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,

\frac{sin(3a)-cos(3a)}{sin(a)+cos(a)} =2sin(2a)-1


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