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

![\frac{(sin(a)+cos(a))[3-4(sin^2(a)+cos^2(a)-sin(a)cos(a)]}{sin(a)+cos(a)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28sin%28a%29%2Bcos%28a%29%29%5B3-4%28sin%5E2%28a%29%2Bcos%5E2%28a%29-sin%28a%29cos%28a%29%5D%7D%7Bsin%28a%29%2Bcos%28a%29%7D%20)
we can use trig identity

![\frac{(sin(a)+cos(a))[3-4(1-sin(a)cos(a)]}{sin(a)+cos(a)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28sin%28a%29%2Bcos%28a%29%29%5B3-4%281-sin%28a%29cos%28a%29%5D%7D%7Bsin%28a%29%2Bcos%28a%29%7D%20)
we can cancel terms

now, we can simplify it further




now, we can use trig identity

we can replace it

so,

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