Prove that sin3a-cos3a/sina+cosa=2sin2a-1
2 answers:
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)
Answer:
Step-by-step explanation:
we are given
we can simplify left side and make it equal to right side
we can use trig identity
now, we can plug values
now, we can simplify
now, we can factor it
we can use trig identity
we can cancel terms
now, we can simplify it further
now, we can use trig identity
we can replace it
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